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#include <bits/stdc++.h>
using namespace std;

struct TestCase {
    int n;
    long long k;
    vector<vector<long long>> A;
    long long answer;
};

mt19937_64 rng_gen(42);

TestCase gen_matrix(int n, long long k, function<long long(int,int)> valfn) {
    TestCase tc;
    tc.n = n; tc.k = k;
    tc.A.assign(n+1, vector<long long>(n+1, 0));
    vector<long long> all;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++) {
            tc.A[i][j] = valfn(i, j);
            all.push_back(tc.A[i][j]);
        }
    sort(all.begin(), all.end());
    tc.answer = all[k-1];
    return tc;
}

TestCase gen_multiplicative(int n, long long k) {
    return gen_matrix(n, k, [](int i, int j) -> long long { return (long long)i * j; });
}
TestCase gen_shifted(int n, long long k) {
    return gen_matrix(n, k, [n](int i, int j) -> long long { return (long long)(i + n) * (j + n); });
}
TestCase gen_additive(int n, long long k) {
    return gen_matrix(n, k, [](int i, int j) -> long long { return i + j; });
}
TestCase gen_random_sorted(int n, long long k) {
    TestCase tc;
    tc.n = n; tc.k = k;
    tc.A.assign(n+1, vector<long long>(n+1, 0));
    long long C = 1000000, D = 1000;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            tc.A[i][j] = (long long)i * C + (long long)j * D + (rng_gen() % 500);
    for (int i = 1; i <= n; i++)
        for (int j = 2; j <= n; j++)
            tc.A[i][j] = max(tc.A[i][j], tc.A[i][j-1]);
    for (int j = 1; j <= n; j++)
        for (int i = 2; i <= n; i++)
            tc.A[i][j] = max(tc.A[i][j], tc.A[i-1][j]);
    vector<long long> all;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            all.push_back(tc.A[i][j]);
    sort(all.begin(), all.end());
    tc.answer = all[k-1];
    return tc;
}

struct Solver {
    const TestCase& tc;
    int query_count;
    vector<long long> memo;
    int n;

    Solver(const TestCase& t) : tc(t), query_count(0), n(t.n) {
        memo.assign(2002 * 2002, -1);
    }

    long long do_query(int r, int c) {
        int key = r * 2001 + c;
        if (memo[key] != -1) return memo[key];
        query_count++;
        memo[key] = tc.A[r][c];
        return memo[key];
    }

    // count_leq with staircase walk, returns count and boundary positions
    long long count_leq(long long x, vector<int>& pos) {
        pos.assign(n + 1, 0);
        long long cnt = 0;
        int j = n;
        for (int i = 1; i <= n; i++) {
            while (j >= 1 && do_query(i, j) > x) j--;
            pos[i] = j;
            cnt += j;
            if (j == 0) break;
        }
        return cnt;
    }

    long long count_lt(long long x, vector<int>& pos) {
        pos.assign(n + 1, 0);
        long long cnt = 0;
        int j = n;
        for (int i = 1; i <= n; i++) {
            while (j >= 1 && do_query(i, j) >= x) j--;
            pos[i] = j;
            cnt += j;
            if (j == 0) break;
        }
        return cnt;
    }

    long long solve() {
        long long k = tc.k;
        long long N2 = (long long)n * n;

        if (n == 1) return do_query(1, 1);
        if (k == 1) return do_query(1, 1);
        if (k == N2) return do_query(n, n);

        long long heap_k = min(k, N2 - k + 1);
        if (heap_k + n <= 24000) {
            if (k <= N2 - k + 1) {
                priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
                for (int i = 1; i <= n; i++) pq.emplace(do_query(i, 1), i, 1);
                long long result = -1;
                for (long long t = 0; t < k; t++) {
                    auto [v, r, c] = pq.top(); pq.pop();
                    result = v;
                    if (c + 1 <= n) pq.emplace(do_query(r, c + 1), r, c + 1);
                }
                return result;
            } else {
                long long kk = N2 - k + 1;
                priority_queue<tuple<long long, int, int>> pq;
                for (int i = 1; i <= n; i++) pq.emplace(do_query(i, n), i, n);
                long long result = -1;
                for (long long t = 0; t < kk; t++) {
                    auto [v, r, c] = pq.top(); pq.pop();
                    result = v;
                    if (c - 1 >= 1) pq.emplace(do_query(r, c - 1), r, c - 1);
                }
                return result;
            }
        }

        // Strategy: query a strategic row, then use binary search on values from that row
        // Row sqrt(k) should contain values near the k-th smallest
        // Actually: query row n/2. The median row has n values spanning roughly
        // from a[n/2][1] to a[n/2][n]. Binary search within this row to find approximate pivot.

        // Better strategy: query the anti-diagonal elements to get initial value range,
        // then do value-based binary search with staircase count_leq

        // Phase 1: Get bounds.
        // Row ceil(k/n) column n is an upper bound for the k-th element.
        int rBound = min(n, (int)((k + n - 1) / n));
        long long hi = do_query(rBound, n);
        long long lo = do_query(1, 1);

        // Verify: count_leq(hi) >= k
        vector<int> posHi;
        long long cntHi = count_leq(hi, posHi);
        if (cntHi < k) {
            hi = do_query(n, n);
            cntHi = count_leq(hi, posHi);
        }

        vector<int> posLo;
        long long cntLo = 0;
        // posLo is all zeros initially
        posLo.assign(n + 1, 0);

        // Phase 2: Narrow via value-based binary search
        // Each count_leq costs O(n) but caches queries
        // Use the row-merge approach: query a strategic row and binary search within it

        // Better: use the "fractional cascading" idea.
        // Query row at index ceil(sqrt(n)) and use values from that row as candidates.
        // Actually, let's use a simpler approach:
        // Binary search: find min value v such that count_leq(v) >= k
        // But we can't binary search on long long values directly (range too large)
        // Instead, generate candidate values by querying strategic positions

        // Strategy: Query entire row at index ~sqrt(k/n) * something
        // Actually the simplest approach that works:
        // Use the quickselect approach but with the staircase from TOP (i ascending, j descending)
        // and use value from the staircase boundary as next pivot

        // Let me implement: quickselect with staircase walk, and use boundary values
        // as pivot candidates for next iteration

        for (int iter = 0; iter < 50; iter++) {
            long long candTotal = cntHi - cntLo;
            if (candTotal <= 0) return hi;

            long long needSmall = k - cntLo;
            long long needLarge = cntHi - k + 1;

            // Count non-empty segments
            int nonempty = 0;
            for (int i = 1; i <= n; i++) {
                if (posHi[i] > posLo[i]) nonempty++;
            }

            long long budget = 49500 - query_count;
            if (min(needSmall, needLarge) + nonempty + 10 <= budget) {
                // Enumerate
                if (needSmall <= needLarge) {
                    priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
                    for (int i = 1; i <= n; i++) {
                        int L = posLo[i] + 1, R = posHi[i];
                        if (L >= 1 && L <= n && L <= R) pq.emplace(do_query(i, L), i, L);
                    }
                    long long result = 0;
                    for (long long t = 0; t < needSmall; t++) {
                        auto [v, r, c] = pq.top(); pq.pop();
                        result = v;
                        if (c + 1 <= posHi[r]) pq.emplace(do_query(r, c + 1), r, c + 1);
                    }
                    return result;
                } else {
                    priority_queue<tuple<long long, int, int>> pq;
                    for (int i = 1; i <= n; i++) {
                        int L = posLo[i] + 1, R = posHi[i];
                        if (R >= 1 && R <= n && L <= R) pq.emplace(do_query(i, R), i, R);
                    }
                    long long result = 0;
                    for (long long t = 0; t < needLarge; t++) {
                        auto [v, r, c] = pq.top(); pq.pop();
                        result = v;
                        if (c - 1 >= posLo[r] + 1) pq.emplace(do_query(r, c - 1), r, c - 1);
                    }
                    return result;
                }
            }

            if (budget < 2 * n + 200) return hi; // fallback

            // Choose pivot: query a strategic position
            // Use the median row's value at the fractional position
            // The "median row" in terms of width-weighted is more complex
            // Simple approach: pick the row with the most candidates and query at target_frac

            // Build prefix sums for position-uniform sampling
            vector<long long> pref(n + 1, 0);
            for (int i = 1; i <= n; i++) {
                int len = posHi[i] - posLo[i];
                pref[i] = pref[i - 1] + max(0, len);
            }
            candTotal = pref[n];
            if (candTotal <= 0) return hi;

            // Sample multiple pivot candidates from the boundary values
            // After a staircase walk for count_leq(v), the cells a[i][posHi[i]] are all <= hi
            // and cells a[i][posHi[i]+1] (if exists) are > hi.
            // The boundary values a[i][posHi[i]] and a[i][posLo[i]+1] are in the cache.
            // Let's use the midpoint of each segment as pivot candidate.

            // Actually, let me try: query the midpoint of each active segment in value terms.
            // For row i, the values range from a[i, posLo[i]+1] to a[i, posHi[i]].
            // The midpoint (in position) is (posLo[i]+1+posHi[i])/2.
            // Sample from several rows, pick weighted quantile.

            double target_q = (double)needSmall / (double)candTotal;

            // Sample from ALL active rows at the target_q position
            vector<pair<long long, long long>> vw; // (value, weight)
            for (int i = 1; i <= n; i++) {
                int L = posLo[i] + 1, R = posHi[i];
                if (L > R || L < 1 || R > n) continue;
                int width = R - L + 1;
                int col = L + (int)(target_q * max(0, width - 1));
                col = max(L, min(R, col));
                vw.push_back({do_query(i, col), width});
            }

            // Weighted quantile
            sort(vw.begin(), vw.end());
            long long target_w = (long long)(target_q * candTotal);
            long long cum = 0;
            long long pivot = vw.back().first;
            for (auto& [v, w] : vw) {
                cum += w;
                if (cum >= target_w) {
                    pivot = v;
                    break;
                }
            }

            // Staircase walk for pivot
            vector<int> posPivot;
            long long cntPivot = count_leq(pivot, posPivot);

            if (cntPivot >= k) {
                if (cntPivot == cntHi && pivot >= hi) {
                    // Pivot didn't help - try count_lt
                    vector<int> posLess;
                    long long cntLess = count_lt(pivot, posLess);
                    if (cntLess < k) return pivot; // answer is pivot
                    hi = pivot - 1; // This is a hack for integer values
                    posHi = posLess;
                    cntHi = cntLess;
                } else {
                    hi = pivot;
                    posHi = posPivot;
                    cntHi = cntPivot;
                }
            } else {
                cntLo = cntPivot;
                posLo = posPivot;
                lo = pivot;
            }
        }
        return -1;
    }
};

int main() {
    struct TestDef {
        string name;
        function<TestCase()> gen;
    };

    vector<TestDef> tests;
    tests.push_back({"additive n=100 k=5000", []{ return gen_additive(100, 5000); }});
    tests.push_back({"multiplicative n=100 k=5000", []{ return gen_multiplicative(100, 5000); }});
    tests.push_back({"additive n=500 k=125000", []{ return gen_additive(500, 125000); }});
    tests.push_back({"multiplicative n=500 k=125000", []{ return gen_multiplicative(500, 125000); }});
    tests.push_back({"random n=500 k=125000", []{ return gen_random_sorted(500, 125000); }});
    tests.push_back({"additive n=2000 k=2000000", []{ return gen_additive(2000, 2000000); }});
    tests.push_back({"multiplicative n=2000 k=2000000", []{ return gen_multiplicative(2000, 2000000); }});
    tests.push_back({"shifted n=2000 k=2000000", []{ return gen_shifted(2000, 2000000); }});
    tests.push_back({"random n=2000 k=2000000", []{ return gen_random_sorted(2000, 2000000); }});
    tests.push_back({"multiplicative n=2000 k=100000", []{ return gen_multiplicative(2000, 100000); }});
    tests.push_back({"multiplicative n=2000 k=3900000", []{ return gen_multiplicative(2000, 3900000); }});

    for (auto& t : tests) {
        auto tc = t.gen();
        Solver s(tc);
        long long result = s.solve();
        bool correct = (result == tc.answer);
        double score;
        int used = s.query_count;
        if (!correct) score = 0.0;
        else if (used <= tc.n) score = 1.0;
        else if (used >= 50000) score = 0.0;
        else score = (50000.0 - used) / (50000.0 - tc.n);

        printf("%-45s n=%4d k=%8lld queries=%6d correct=%s score=%.4f\n",
               t.name.c_str(), tc.n, tc.k, used, correct ? "YES" : "NO", score);
    }
    return 0;
}

	#include <bits/stdc++.h>
	using namespace std;

	struct TestCase {
	int n;
	long long k;
	vector<vector<long long>> A;
	long long answer;
	};

	mt19937_64 rng_gen(42);

	TestCase gen_matrix(int n, long long k, function<long long(int,int)> valfn) {
	TestCase tc;
	tc.n = n; tc.k = k;
	tc.A.assign(n+1, vector<long long>(n+1, 0));
	vector<long long> all;
	for (int i = 1; i <= n; i++)
	for (int j = 1; j <= n; j++) {
	tc.A[i][j] = valfn(i, j);
	all.push_back(tc.A[i][j]);
	}
	sort(all.begin(), all.end());
	tc.answer = all[k-1];
	return tc;
	}

	TestCase gen_multiplicative(int n, long long k) {
	return gen_matrix(n, k, [](int i, int j) -> long long { return (long long)i * j; });
	}
	TestCase gen_shifted(int n, long long k) {
	return gen_matrix(n, k, [n](int i, int j) -> long long { return (long long)(i + n) * (j + n); });
	}
	TestCase gen_additive(int n, long long k) {
	return gen_matrix(n, k, [](int i, int j) -> long long { return i + j; });
	}
	TestCase gen_random_sorted(int n, long long k) {
	TestCase tc;
	tc.n = n; tc.k = k;
	tc.A.assign(n+1, vector<long long>(n+1, 0));
	long long C = 1000000, D = 1000;
	for (int i = 1; i <= n; i++)
	for (int j = 1; j <= n; j++)
	tc.A[i][j] = (long long)i * C + (long long)j * D + (rng_gen() % 500);
	for (int i = 1; i <= n; i++)
	for (int j = 2; j <= n; j++)
	tc.A[i][j] = max(tc.A[i][j], tc.A[i][j-1]);
	for (int j = 1; j <= n; j++)
	for (int i = 2; i <= n; i++)
	tc.A[i][j] = max(tc.A[i][j], tc.A[i-1][j]);
	vector<long long> all;
	for (int i = 1; i <= n; i++)
	for (int j = 1; j <= n; j++)
	all.push_back(tc.A[i][j]);
	sort(all.begin(), all.end());
	tc.answer = all[k-1];
	return tc;
	}

	struct Solver {
	const TestCase& tc;
	int query_count;
	vector<long long> memo;
	int n;

	Solver(const TestCase& t) : tc(t), query_count(0), n(t.n) {
	memo.assign(2002 * 2002, -1);
	}

	long long do_query(int r, int c) {
	int key = r * 2001 + c;
	if (memo[key] != -1) return memo[key];
	query_count++;
	memo[key] = tc.A[r][c];
	return memo[key];
	}

	// count_leq with staircase walk, returns count and boundary positions
	long long count_leq(long long x, vector<int>& pos) {
	pos.assign(n + 1, 0);
	long long cnt = 0;
	int j = n;
	for (int i = 1; i <= n; i++) {
	while (j >= 1 && do_query(i, j) > x) j--;
	pos[i] = j;
	cnt += j;
	if (j == 0) break;
	}
	return cnt;
	}

	long long count_lt(long long x, vector<int>& pos) {
	pos.assign(n + 1, 0);
	long long cnt = 0;
	int j = n;
	for (int i = 1; i <= n; i++) {
	while (j >= 1 && do_query(i, j) >= x) j--;
	pos[i] = j;
	cnt += j;
	if (j == 0) break;
	}
	return cnt;
	}

	long long solve() {
	long long k = tc.k;
	long long N2 = (long long)n * n;

	if (n == 1) return do_query(1, 1);
	if (k == 1) return do_query(1, 1);
	if (k == N2) return do_query(n, n);

	long long heap_k = min(k, N2 - k + 1);
	if (heap_k + n <= 24000) {
	if (k <= N2 - k + 1) {
	priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
	for (int i = 1; i <= n; i++) pq.emplace(do_query(i, 1), i, 1);
	long long result = -1;
	for (long long t = 0; t < k; t++) {
	auto [v, r, c] = pq.top(); pq.pop();
	result = v;
	if (c + 1 <= n) pq.emplace(do_query(r, c + 1), r, c + 1);
	}
	return result;
	} else {
	long long kk = N2 - k + 1;
	priority_queue<tuple<long long, int, int>> pq;
	for (int i = 1; i <= n; i++) pq.emplace(do_query(i, n), i, n);
	long long result = -1;
	for (long long t = 0; t < kk; t++) {
	auto [v, r, c] = pq.top(); pq.pop();
	result = v;
	if (c - 1 >= 1) pq.emplace(do_query(r, c - 1), r, c - 1);
	}
	return result;
	}
	}

	// Strategy: query a strategic row, then use binary search on values from that row
	// Row sqrt(k) should contain values near the k-th smallest
	// Actually: query row n/2. The median row has n values spanning roughly
	// from a[n/2][1] to a[n/2][n]. Binary search within this row to find approximate pivot.

	// Better strategy: query the anti-diagonal elements to get initial value range,
	// then do value-based binary search with staircase count_leq

	// Phase 1: Get bounds.
	// Row ceil(k/n) column n is an upper bound for the k-th element.
	int rBound = min(n, (int)((k + n - 1) / n));
	long long hi = do_query(rBound, n);
	long long lo = do_query(1, 1);

	// Verify: count_leq(hi) >= k
	vector<int> posHi;
	long long cntHi = count_leq(hi, posHi);
	if (cntHi < k) {
	hi = do_query(n, n);
	cntHi = count_leq(hi, posHi);
	}

	vector<int> posLo;
	long long cntLo = 0;
	// posLo is all zeros initially
	posLo.assign(n + 1, 0);

	// Phase 2: Narrow via value-based binary search
	// Each count_leq costs O(n) but caches queries
	// Use the row-merge approach: query a strategic row and binary search within it

	// Better: use the "fractional cascading" idea.
	// Query row at index ceil(sqrt(n)) and use values from that row as candidates.
	// Actually, let's use a simpler approach:
	// Binary search: find min value v such that count_leq(v) >= k
	// But we can't binary search on long long values directly (range too large)
	// Instead, generate candidate values by querying strategic positions

	// Strategy: Query entire row at index ~sqrt(k/n) * something
	// Actually the simplest approach that works:
	// Use the quickselect approach but with the staircase from TOP (i ascending, j descending)
	// and use value from the staircase boundary as next pivot

	// Let me implement: quickselect with staircase walk, and use boundary values
	// as pivot candidates for next iteration

	for (int iter = 0; iter < 50; iter++) {
	long long candTotal = cntHi - cntLo;
	if (candTotal <= 0) return hi;

	long long needSmall = k - cntLo;
	long long needLarge = cntHi - k + 1;

	// Count non-empty segments
	int nonempty = 0;
	for (int i = 1; i <= n; i++) {
	if (posHi[i] > posLo[i]) nonempty++;
	}

	long long budget = 49500 - query_count;
	if (min(needSmall, needLarge) + nonempty + 10 <= budget) {
	// Enumerate
	if (needSmall <= needLarge) {
	priority_queue<tuple<long long, int, int>, vector<tuple<long long, int, int>>, greater<>> pq;
	for (int i = 1; i <= n; i++) {
	int L = posLo[i] + 1, R = posHi[i];
	if (L >= 1 && L <= n && L <= R) pq.emplace(do_query(i, L), i, L);
	}
	long long result = 0;
	for (long long t = 0; t < needSmall; t++) {
	auto [v, r, c] = pq.top(); pq.pop();
	result = v;
	if (c + 1 <= posHi[r]) pq.emplace(do_query(r, c + 1), r, c + 1);
	}
	return result;
	} else {
	priority_queue<tuple<long long, int, int>> pq;
	for (int i = 1; i <= n; i++) {
	int L = posLo[i] + 1, R = posHi[i];
	if (R >= 1 && R <= n && L <= R) pq.emplace(do_query(i, R), i, R);
	}
	long long result = 0;
	for (long long t = 0; t < needLarge; t++) {
	auto [v, r, c] = pq.top(); pq.pop();
	result = v;
	if (c - 1 >= posLo[r] + 1) pq.emplace(do_query(r, c - 1), r, c - 1);
	}
	return result;
	}
	}

	if (budget < 2 * n + 200) return hi; // fallback

	// Choose pivot: query a strategic position
	// Use the median row's value at the fractional position
	// The "median row" in terms of width-weighted is more complex
	// Simple approach: pick the row with the most candidates and query at target_frac

	// Build prefix sums for position-uniform sampling
	vector<long long> pref(n + 1, 0);
	for (int i = 1; i <= n; i++) {
	int len = posHi[i] - posLo[i];
	pref[i] = pref[i - 1] + max(0, len);
	}
	candTotal = pref[n];
	if (candTotal <= 0) return hi;

	// Sample multiple pivot candidates from the boundary values
	// After a staircase walk for count_leq(v), the cells a[i][posHi[i]] are all <= hi
	// and cells a[i][posHi[i]+1] (if exists) are > hi.
	// The boundary values a[i][posHi[i]] and a[i][posLo[i]+1] are in the cache.
	// Let's use the midpoint of each segment as pivot candidate.

	// Actually, let me try: query the midpoint of each active segment in value terms.
	// For row i, the values range from a[i, posLo[i]+1] to a[i, posHi[i]].
	// The midpoint (in position) is (posLo[i]+1+posHi[i])/2.
	// Sample from several rows, pick weighted quantile.

	double target_q = (double)needSmall / (double)candTotal;

	// Sample from ALL active rows at the target_q position
	vector<pair<long long, long long>> vw; // (value, weight)
	for (int i = 1; i <= n; i++) {
	int L = posLo[i] + 1, R = posHi[i];
	if (L > R \|\| L < 1 \|\| R > n) continue;
	int width = R - L + 1;
	int col = L + (int)(target_q * max(0, width - 1));
	col = max(L, min(R, col));
	vw.push_back({do_query(i, col), width});
	}

	// Weighted quantile
	sort(vw.begin(), vw.end());
	long long target_w = (long long)(target_q * candTotal);
	long long cum = 0;
	long long pivot = vw.back().first;
	for (auto& [v, w] : vw) {
	cum += w;
	if (cum >= target_w) {
	pivot = v;
	break;
	}
	}

	// Staircase walk for pivot
	vector<int> posPivot;
	long long cntPivot = count_leq(pivot, posPivot);

	if (cntPivot >= k) {
	if (cntPivot == cntHi && pivot >= hi) {
	// Pivot didn't help - try count_lt
	vector<int> posLess;
	long long cntLess = count_lt(pivot, posLess);
	if (cntLess < k) return pivot; // answer is pivot
	hi = pivot - 1; // This is a hack for integer values
	posHi = posLess;
	cntHi = cntLess;
	} else {
	hi = pivot;
	posHi = posPivot;
	cntHi = cntPivot;
	}
	} else {
	cntLo = cntPivot;
	posLo = posPivot;
	lo = pivot;
	}
	}
	return -1;
	}
	};

	int main() {
	struct TestDef {
	string name;
	function<TestCase()> gen;
	};

	vector<TestDef> tests;
	tests.push_back({"additive n=100 k=5000", []{ return gen_additive(100, 5000); }});
	tests.push_back({"multiplicative n=100 k=5000", []{ return gen_multiplicative(100, 5000); }});
	tests.push_back({"additive n=500 k=125000", []{ return gen_additive(500, 125000); }});
	tests.push_back({"multiplicative n=500 k=125000", []{ return gen_multiplicative(500, 125000); }});
	tests.push_back({"random n=500 k=125000", []{ return gen_random_sorted(500, 125000); }});
	tests.push_back({"additive n=2000 k=2000000", []{ return gen_additive(2000, 2000000); }});
	tests.push_back({"multiplicative n=2000 k=2000000", []{ return gen_multiplicative(2000, 2000000); }});
	tests.push_back({"shifted n=2000 k=2000000", []{ return gen_shifted(2000, 2000000); }});
	tests.push_back({"random n=2000 k=2000000", []{ return gen_random_sorted(2000, 2000000); }});
	tests.push_back({"multiplicative n=2000 k=100000", []{ return gen_multiplicative(2000, 100000); }});
	tests.push_back({"multiplicative n=2000 k=3900000", []{ return gen_multiplicative(2000, 3900000); }});

	for (auto& t : tests) {
	auto tc = t.gen();
	Solver s(tc);
	long long result = s.solve();
	bool correct = (result == tc.answer);
	double score;
	int used = s.query_count;
	if (!correct) score = 0.0;
	else if (used <= tc.n) score = 1.0;
	else if (used >= 50000) score = 0.0;
	else score = (50000.0 - used) / (50000.0 - tc.n);

	printf("%-45s n=%4d k=%8lld queries=%6d correct=%s score=%.4f\n",
	t.name.c_str(), tc.n, tc.k, used, correct ? "YES" : "NO", score);
	}
	return 0;
	}