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;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long MOD = 1000000007;
  const char nl = '\n';
  const long long MX = 100001;
  long long powi(long long x, long long n) {
    long long res = 1;
    while (n) {
      if (n & 1) res = (res * x) % MOD;
      x = (x * x) % MOD;
      n >>= 1;
    }
    return res;
  }
  bool isPowerOfTwo(long long n) {
    if (n == 0) return 0;
    while (n != 1) {
      if (n % 2 != 0) return 0;
      n = n / 2;
    }
    return 1;
  }
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    a %= b;
    return gcd(b, a);
  }
  vector<long long> second[100005];
  bool check(long long p, vector<vector<long long>>& matrix, long long n,
             long long m) {
    for (long long i = 0; i < (n); i++) {
      for (long long j = 0; j < (m); j++) {
        if (p <= matrix[j][i]) {
          second[i].push_back(j);
        }
      }
    }
    bool first = 0;
    long long cnt = 0;
    for (long long j = 0; j < (m); j++) {
      for (long long i = 0; i < (n); i++) {
        if (second[i].empty()) {
          cnt = 0;
          break;
        }
        bool f2 = 0;
        for (auto u : second[i]) {
          if (u == j) {
            f2 = 1;
            break;
          }
        }
        if (f2 == 1) {
          cnt++;
        }
      }
      if (cnt > 1) return true;
      cnt = 0;
    }
    return false;
  }
  void solve() {
    long long m, n;
    cin >> m >> n;
    vector<vector<long long>> matrix(m, vector<long long>(n));
    for (long long i = 0; i < (m); i++) {
      for (long long j = 0; j < (n); j++) cin >> matrix[i][j];
    }
    if (n - 1 >= m) {
      long long mini = 1;
      long long cmini = 1e18;
      for (long long j = 0; j < (n); j++) {
        mini = 1;
        for (long long i = 0; i < (m); i++) {
          mini = max(mini, matrix[i][j]);
        }
        cmini = min(mini, cmini);
      }
      cout << cmini << "\n";
      return;
    }
    long long st = 1;
    long long end = 1e10;
    long long ans = 1;
    while (st <= end) {
      for (long long i = 0; i < (n + 1); i++) second[i].clear();
      long long mid = st + (end - st) / 2;
      if (check(mid, matrix, n, m)) {
        st = mid + 1;
        ans = max(ans, mid);
      } else {
        end = mid - 1;
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  