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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  void fileio(const string &s) {
    freopen((s + ".in").c_str(), "r", stdin);
    freopen((s + ".out").c_str(), "w", stdout);
  }
  const long long INF = 4e18;
  inline long long read() {
    long long x = 0;
    bool flag = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
      if (c == '-') flag = 0;
      c = getchar();
    }
    while (c >= '0' && c <= '9') {
      x = (x << 1) + (x << 3) + c - '0';
      c = getchar();
    }
    return (flag ? x : ~(x - 1));
  }
  long long t, n, m, vis[100000], mst[100000], num[100000];
  vector<long long> p[100000], ok[100000];
  bool check(long long x) {
    fill(vis, vis + m, 0);
    fill(mst, mst + n, 0);
    fill(num, num + n, 0);
    for (long long i = 0; i < m; i++) ok[i].clear();
    for (long long i = 0; i < m; i++)
      for (long long l = 0; l < n; l++) ok[i].push_back(0);
    for (long long i = 0; i < n; i++)
      for (long long l = 0; l < m; l++)
        if (p[i][l] >= x) {
          ok[l][i] = 1;
          vis[l]++;
          num[i]++;
        }
    for (long long i = 0; i < m; i++)
      if (!vis[i]) return 0;
    for (long long i = 0; i < m; i++)
      for (long long l = 0; l < n; l++)
        if (vis[i] == 1 && ok[i][l]) mst[l] = 1;
    long long res = 0;
    for (long long i = 0; i < n; i++) res += mst[i];
    if (res >= m) return 0;
    for (long long i = 0; i < n; i++)
      if (num[i] > 1) return 1;
    return 0;
  }
  void solve() {
    n = read();
    m = read();
    for (long long i = 0; i < n; i++) {
      p[i].clear();
      for (long long l = 1; l <= m; l++) p[i].push_back(read());
    }
    long long l = 1, r = 1000000000;
    while (l <= r) {
      long long mid = (l + r) >> 1;
      if (check(mid))
        l = mid + 1;
      else
        r = mid - 1;
    }
    cout << l - 1 << endl;
  }
  signed main() {
    t = read();
    while (t--) solve();
    return 0;
  }
  