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 gcd(long long x, long long y) {
    if (x == 0) return y;
    return (gcd(y % x, x));
  }
  string Tolower(string s) {
    transform(s.begin(), s.end(), s.begin(), ::tolower);
    return s;
  }
  string Toupper(string s) {
    transform(s.begin(), s.end(), s.begin(), ::toupper);
    return s;
  }
  void solve() {
    long long t;
    cin >> t;
    while (t--) {
      long long m, n, ans = -1, l, r, mid;
      cin >> m >> n;
      vector<vector<long long>> a(m);
      set<long long> s;
      vector<long long> b;
      for (long long i = 0; i < m; i++) {
        for (long long j = 0; j < n; j++) {
          long long x;
          cin >> x;
          a[i].push_back(x);
          s.insert(x);
        }
      }
      for (auto x : s) b.push_back(x);
      l = 0, r = b.size() - 1;
      while (l <= r) {
        mid = l + (r - l) / 2;
        bool f = true;
        map<long long, long long> make_pair;
        set<long long> sex;
        for (long long i = 0; i < m; i++) {
          for (long long j = 0; j < n; j++) {
            if (a[i][j] >= b[mid]) {
              make_pair[i]++;
              sex.insert(j);
            }
          }
        }
        vector<long long> temp;
        for (auto x : make_pair) temp.push_back(x.second);
        sort((temp).begin(), (temp).end(), greater<int>());
        long long sum = 0;
        long long sz = temp.size();
        for (long long i = 0; i < min(n - 1, sz); i++) {
          sum += temp[i];
        }
        if (sum >= n && (long long)sex.size() == n)
          f = true;
        else
          f = false;
        if (f) {
          ans = b[mid];
          l = mid + 1;
        } else
          r = mid - 1;
      }
      if (n - 1 >= m) {
        ans = 1e9;
        for (long long i = 0; i < n; i++) {
          long long mx = -1;
          for (long long j = 0; j < m; j++) {
            mx = max(mx, a[j][i]);
          }
          ans = min(ans, mx);
        }
      }
      cout << ans << "\n";
    }
    return;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
    return 0;
  }
  