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;
  #pragma GCC target("avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  void solve() {
    long long m, n, mx = 0, mn = 1e10;
    cin >> m >> n;
    vector<vector<long long>> grid(m, vector<long long>(n));
    for (long long i = 0; i < m; i++)
      for (long long j = 0; j < n; j++) {
        cin >> grid[i][j];
        mx = max(mx, grid[i][j]);
        mn = min(mn, grid[i][j]);
      }
    auto isPossible = [&](long long val) {
      vector<set<long long>> pos(m);
      for (long long i = 0; i < m; i++)
        for (long long j = 0; j < n; j++) {
          if (grid[i][j] >= val) {
            pos[i].insert(j);
          }
        }
      set<long long> st, idx;
      long long cnt = 0;
      bool flag = true;
      while (st.size() < n) {
        long long maxsize = 0, maxid = -1;
        for (long long i = 0; i < m; i++) {
          if (pos[i].size() > maxsize) {
            maxsize = pos[i].size();
            maxid = i;
          }
        }
        if (maxid == -1) {
          flag = false;
          break;
        } else {
          idx.insert(maxid);
          for (auto &x : pos[maxid]) {
            st.insert(x);
          }
          for (long long i = 0; i < m; i++) {
            vector<long long> del;
            for (auto &x : pos[i]) {
              if (st.find(x) != st.end()) {
                del.push_back(x);
              }
            }
            for (auto &x : del) pos[i].erase(x);
          }
        }
      }
      return (idx.size() <= (n - 1) && flag);
    };
    long long s = mn, e = mx, res = s;
    while (s <= e) {
      long long mid = s + (e - s) / 2;
      if (isPossible(mid)) {
        res = mid;
        s = mid + 1;
      } else {
        e = mid - 1;
      }
    }
    cout << res << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  