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 <typename A, typename B>
  string to_string(pair<A, B> p);
  template <typename A, typename B, typename C>
  string to_string(tuple<A, B, C> p);
  template <typename A, typename B, typename C, typename D>
  string to_string(tuple<A, B, C, D> p);
  string to_string(const string& s) { return '"' + s + '"'; }
  string to_string(const char* s) { return to_string((string)s); }
  string to_string(bool b) { return (b ? "true" : "false"); }
  string to_string(vector<bool> v) {
    bool first = true;
    string res = "{";
    for (int i = 0; i < static_cast<int>(v.size()); i++) {
      if (!first) {
        res += ", ";
      }
      first = false;
      res += to_string(v[i]);
    }
    res += "}";
    return res;
  }
  template <size_t N>
  string to_string(bitset<N> v) {
    string res = "";
    for (size_t i = 0; i < N; i++) {
      res += static_cast<char>('0' + v[i]);
    }
    return res;
  }
  template <typename A>
  string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto& x : v) {
      if (!first) {
        res += ", ";
      }
      first = false;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  template <typename A, typename B>
  string to_string(pair<A, B> p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <typename A, typename B, typename C>
  string to_string(tuple<A, B, C> p) {
    return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
           to_string(get<2>(p)) + ")";
  }
  template <typename A, typename B, typename C, typename D>
  string to_string(tuple<A, B, C, D> p) {
    return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
           to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
  }
  void debug_out() { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  int main() {
    ios::sync_with_stdio(false);
    int T;
    { cin >> T; }
    for (int kase = 1; kase <= T; ++kase) {
      int M, N;
      { cin >> M >> N; }
      vector<vector<int>> stores(M, vector<int>(N));
      {
        for (int i = 0; i < M; ++i) {
          for (int j = 0; j < N; ++j) {
            cin >> stores[i][j];
          }
        }
      }
      int ans = 0;
      {
        vector<int> f(N);
        for (int i = 0; i < M; ++i) {
          for (int j = 0; j < N; ++j) {
            f[j] = max(f[j], stores[i][j]);
          }
        }
        if (N - 1 >= M) {
          ans = *min_element(f.begin(), f.end());
        } else {
          for (int i = 0; i < N; ++i) {
            for (int j = i + 1; j < N; ++j) {
              int fij = 0;
              {
                for (int k = 0; k < M; ++k) {
                  fij = max(fij, min(stores[k][i], stores[k][j]));
                }
              }
              for (int k = 0; k < N; ++k)
                if (k != i && k != j) fij = min(fij, f[k]);
              ans = max(ans, fij);
            }
          }
        }
      }
      cout << ans << "\n";
    }
  }
  