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;
  const long long INF = 1e9;
  const long long INFF = 1e18;
  const long long MAXN = 510;
  const long long MOD = 1e9 + 7;
  const double PI = acos(-1.0);
  const double INFD = 1E9;
  const double EPS = 1e-9;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long getTC() {
    if (1) {
      long long t;
      cin >> t;
      return t;
    } else
      return 1;
  }
  void solve() {
    long long n, m;
    cin >> m >> n;
    vector<vector<long long>> tab(m, vector<long long>(n));
    for (auto &it : tab)
      for (auto &it2 : it) cin >> it2;
    const auto cek = [&](long long x) -> bool {
      bool ada = 0;
      for (long long i = 0; i < m; i++) {
        long long cnt = 0;
        for (long long j = 0; j < n; j++) {
          cnt += (tab[i][j] >= x);
        }
        ada |= (cnt > 1);
      }
      bool valid = 1;
      for (long long j = 0; j < n; j++) {
        bool hehe = 0;
        for (long long i = 0; i < m; i++) {
          hehe |= (tab[i][j] >= x);
        }
        valid &= hehe;
      }
      return (ada && valid);
    };
    long long l = 1;
    long long r = INF;
    while (r - l > 1) {
      long long mid = (l + r) >> 1;
      if (cek(mid))
        l = mid;
      else
        r = mid - 1;
    }
    if (cek(r))
      cout << r << endl;
    else
      cout << l << endl;
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t = getTC();
    while (t--) {
      solve();
    }
    return 0;
  }
  