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 int md = 1000000007;
  long long int add(long long int a, long long int b) {
    return (a % md + b % md + md) % md;
  }
  long long int sub(long long int a, long long int b) {
    return (a % md + md - b % md) % md;
  }
  long long int mul(long long int a, long long int b) {
    return ((a % md) * (b % md) + md) % md;
  }
  const long long int INF = 1e18;
  long long int dirA[] = {0, -1, 0, 1};
  long long int dirB[] = {-1, 0, 1, 0};
  long long int dirX[] = {0, -1, 0, 1, -1, -1, 1, 1};
  long long int dirY[] = {-1, 0, 1, 0, -1, 1, -1, 1};
  long long int power(long long int a, long long int b, long long int p) {
    if (a == 0) return 0;
    long long int res = 1;
    a %= p;
    while (b > 0) {
      if (b & 1) res = (res * a) % p;
      b >>= 1;
      a = (a * a) % p;
    }
    return res;
  }
  long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
  long long int min(long long int a, long long int b) { return (a < b) ? a : b; }
  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 / gcd(a, b)) * b;
  }
  vector<bool> is_prime(100000001, true);
  vector<long long int> primes;
  void isPrime() {
    long long int n = 1000000;
    is_prime[0] = is_prime[1] = false;
    for (long long int i = 2; i <= n; i++) {
      if (is_prime[i] && i * i <= n) {
        for (long long int j = i * i; j <= n; j += i) {
          is_prime[j] = false;
        }
      }
    }
  }
  bool sortbysec(const pair<long long int, long long int>& a,
                 const pair<long long int, long long int>& b) {
    return (a.second < b.second);
  }
  vector<long long int> two;
  void solve() {
    long long int m, n;
    cin >> m >> n;
    vector<vector<long long int> > joy(m, vector<long long int>(n, 0));
    for (long long int i = 0; i < m; i++) {
      for (long long int j = 0; j < n; j++) {
        cin >> joy[i][j];
      }
    }
    long long int lo = 0, hi = INF, mid = 0;
    long long int ans = 0;
    while (lo <= hi) {
      mid = lo + (hi - lo) / 2;
      set<long long int> s;
      bool valid = false, shops = false;
      for (long long int i = 0; i < n; i++) {
        valid = false;
        for (long long int j = 0; j < m; j++) {
          if (joy[j][i] >= mid) {
            valid = true;
            if (s.count(j)) {
              shops = true;
            }
            s.insert(j);
          }
        }
        if (!valid) break;
      }
      if (valid && shops) {
        ans = mid;
        lo = mid + 1;
      } else {
        hi = mid - 1;
      }
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ;
    isPrime();
    long long int t = 1;
    cin >> t;
    long long int a = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  