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;
  bool isPrime(long long int n) {
    if (n <= 1) return false;
    if (n <= 3) return true;
    if (n % 2 == 0 || n % 3 == 0) return false;
    long long int p = sqrt(n);
    for (int i = 5; i <= p; i += 6)
      if (n % i == 0 || n % (i + 2) == 0) return false;
    return true;
  }
  string d2b(int n) {
    string s;
    for (int i = 31; i >= 0; i--) {
      int k = n >> i;
      if (k & 1)
        s += '1';
      else
        s += "0";
    }
    return s;
  }
  long long int gcd(long long int a, long long int b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long int lcm(long long int a, long long int b) {
    return (a * b) / gcd(a, b);
  }
  long long int b2d(string n) {
    string num = n;
    long long int dec_value = 0;
    long long int base = 1;
    long long int len = num.length();
    for (int i = len - 1; i >= 0; i--) {
      if (num[i] == '1') dec_value += base;
      base = base * 2;
    }
    return dec_value;
  }
  bool check(long long int x, vector<vector<long long int>> &v) {
    long long int m = v.size();
    long long int n = v[0].size();
    unordered_map<long long int, long long int> mp;
    for (long long j = 0; j < n; j++) {
      bool f = 0;
      for (long long i = 0; i < m; i++) {
        if (v[i][j] >= x) {
          f = 1;
          break;
        }
      }
      if (f == 0) return 0;
    }
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        if (v[i][j] >= x) {
          if (mp[i] == 1) {
            return 1;
          }
          mp[i] = 1;
        }
      }
    }
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int test_case = 1;
    cin >> test_case;
    while (test_case--) {
      long long int n, m;
      cin >> m >> n;
      vector<vector<long long int>> a(m, vector<long long int>(n));
      long long int l = INT_MAX, h = INT_MIN;
      for (long long i = 0; i < m; i++) {
        for (long long j = 0; j < n; j++) {
          cin >> a[i][j];
          l = min(l, a[i][j]);
          h = max(h, a[i][j]);
        }
      }
      long long int mid;
      long long int ans = 0;
      while (l <= h) {
        mid = (l + h) / 2;
        if (check(mid, a)) {
          ans = mid;
          l = mid + 1;
        } else
          h = mid - 1;
      }
      cout << ans << endl;
    }
  }
  