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;
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  mt19937_64 rang(
      chrono::high_resolution_clock::now().time_since_epoch().count());
  int rng(int lim) {
    uniform_int_distribution<int> uid(0, lim - 1);
    return uid(rang);
  }
  int mpow(int base, int exp);
  void ipgraph(int n, int m);
  void dfs(int u, int par);
  const int mod = 1000000007;
  const int N = 3e5, M = N;
  vector<int> g[N];
  int a[N];
  void solve() {
    int i, j, n, m;
    cin >> m >> n;
    vector<vector<int> > vec(m, vector<int>(n));
    vector<int> mx(n, INT_MIN);
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        cin >> vec[i][j];
      }
    }
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        mx[i] = max(mx[i], vec[j][i]);
      }
    }
    for (i = 0; i < m; i++) {
      sort(vec[i].begin(), vec[i].end());
    }
    int maxi = INT_MIN;
    for (i = 0; i < m; i++) {
      maxi = max(maxi, vec[i][n - 2]);
    }
    int ans = INT_MAX;
    for (i = 0; i < n; i++) {
      ans = min(ans, mx[i]);
    }
    ans = min(ans, maxi);
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
    int t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  int mpow(int base, int exp) {
    base %= mod;
    int result = 1;
    while (exp > 0) {
      if (exp & 1) result = ((long long)result * base) % mod;
      base = ((long long)base * base) % mod;
      exp >>= 1;
    }
    return result;
  }
  void ipgraph(int n, int m) {
    int i, u, v;
    while (m--) {
      cin >> u >> v;
      u--, v--;
      g[u].push_back(v);
      g[v].push_back(u);
    }
  }
  void dfs(int u, int par) {
    for (int v : g[u]) {
      if (v == par) continue;
      dfs(v, u);
    }
  }
  