description: Vlad has n friends, for each of whom he wants to buy one gift for the New Year.
  
  There are m shops in the city, in each of which he can buy a gift for any of his friends. If the j-th friend (1 ≤ j ≤ n) receives a gift bought in the shop with the number i (1 ≤ i ≤ m), then the friend receives p_{ij} units of joy. The rectangular table p_{ij} is given in the input.
  
  Vlad has time to visit at most n-1 shops (where n is the number of friends). He chooses which shops he will visit and for which friends he will buy gifts in each of them.
  
  Let the j-th friend receive a_j units of joy from Vlad's gift. Let's find the value α=min\\{a_1, a_2, ..., a_n\}. Vlad's goal is to buy gifts so that the value of α is as large as possible. In other words, Vlad wants to maximize the minimum of the joys of his friends.
  
  For example, let m = 2, n = 2. Let the joy from the gifts that we can buy in the first shop: p_{11} = 1, p_{12}=2, in the second shop: p_{21} = 3, p_{22}=4.
  
  Then it is enough for Vlad to go only to the second shop and buy a gift for the first friend, bringing joy 3, and for the second — bringing joy 4. In this case, the value α will be equal to min\{3, 4\} = 3
  
  Help Vlad choose gifts for his friends so that the value of α is as high as possible. Please note that each friend must receive one gift. Vlad can visit at most n-1 shops (where n is the number of friends). In the shop, he can buy any number of gifts.
  
  Input
  
  The first line of the input contains an integer t (1 ≤ t ≤ 10^4) — the number of test cases in the input.
  
  An empty line is written before each test case. Then there is a line containing integers m and n (2 ≤ n, 2 ≤ n ⋅ m ≤ 10^5) separated by a space — the number of shops and the number of friends, where n ⋅ m is the product of n and m.
  
  Then m lines follow, each containing n numbers. The number in the i-th row of the j-th column p_{ij} (1 ≤ p_{ij} ≤ 10^9) is the joy of the product intended for friend number j in shop number i.
  
  It is guaranteed that the sum of the values n ⋅ m over all test cases in the test does not exceed 10^5.
  
  Output
  
  Print t lines, each line must contain the answer to the corresponding test case — the maximum possible value of α, where α is the minimum of the joys from a gift for all of Vlad's friends.
  
  Example
  
  Input
  
  
  5
  
  2 2
  1 2
  3 4
  
  4 3
  1 3 1
  3 1 1
  1 2 2
  1 1 3
  
  2 3
  5 3 4
  2 5 1
  
  4 2
  7 9
  8 1
  9 6
  10 8
  
  2 4
  6 5 2 1
  7 9 7 2
  
  
  Output
  
  
  3
  2
  4
  8
  2
solution: #include <bits/stdc++.h>
  using namespace std;
  long long t = 1, n, m, k, q, l, r, a[100009], b, c, u, wtf[100009], x, y, z,
            mod = 1e9 + 7;
  string s;
  vector<long long> v[100009];
  struct st {
    long long x, y;
  };
  bool cmp(const st &a, const st &b) { return a.x < b.x; }
  inline long long read() {
    long long s = 0, w = 1;
    char ch = getchar();
    while (ch > '9' || ch < '0') {
      if (ch == '-') w = -1;
      ch = getchar();
    }
    while (ch <= '9' && ch >= '0')
      s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return s * w;
  }
  long long qp(long long x, long long y) {
    long long a = 1, b = x;
    while (y) {
      if (y & 1) a = a * b % mod;
      b = b * b % mod, y >>= 1;
    }
    return a;
  }
  bool judge(long long x) {
    fill(wtf + 1, wtf + m + 1, 0);
    bool qwq = 0;
    for (long long i = 1; i <= n; i += 1) {
      bool fl = 0;
      for (long long j = 1; j <= m; j += 1) {
        if (v[j][i] >= x && wtf[j]) qwq = 1;
        if (v[j][i] >= x) wtf[j] = 1, fl = 1;
      }
      if (!fl) return 0;
    }
    return qwq;
  }
  int main() {
    t = read();
    while (t--) {
      m = read(), n = read();
      fill(a + 1, a + n + 1, 0);
      for (long long i = 1; i <= m; i += 1) v[i].resize(n + 2);
      for (long long i = 1; i <= m; i += 1)
        for (long long j = 1; j <= n; j += 1) v[i][j] = read();
      if (n > m) {
        for (long long i = 1; i <= m; i += 1)
          for (long long j = 1; j <= n; j += 1) a[j] = max(a[j], v[i][j]);
        printf("%lld\n", *min_element(a + 1, a + n + 1));
      } else {
        long long l = 1, r = 1e9, ans = 0;
        while (l <= r) {
          long long mid = (l + r) >> 1;
          if (judge(mid))
            ans = mid, l = mid + 1;
          else
            r = mid - 1;
        }
        printf("%lld\n", ans);
      }
    }
    return 0;
  }
  