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;
  #pragma GCC target("avx2")
  #pragma GCC optimization("O3")
  #pragma GCC optimization("unroll-loops")
  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(char x) { return (string) "\'" + x + "\'"; }
  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 (long long i = 0; i < static_cast<long long>(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 << "]\n"; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << to_string(H) << (sizeof...(T) != 0 ? ", " : "");
    debug_out(T...);
  }
  template <typename T>
  void debug_out(T A[], long long SZ) {
    cerr << "{";
    for (long long i = 0; i < SZ; i++) cerr << A[i] << (i != SZ - 1 ? ", " : "");
    cerr << "}]\n";
  }
  const long long INF = (long long)2e18;
  void solve() {
    long long m, n;
    cin >> m >> n;
    vector<vector<long long>> grid(m, vector<long long>(n));
    for (long long i = 0; i < m; i++)
      for (long long j = 0; j < n; j++) cin >> grid[i][j];
    auto isPossible = [&](long long val) {
      long long flag1 = true, flag2 = false;
      set<long long> st;
      for (long long j = 0; j < n; j++) {
        bool possible = false;
        for (long long i = 0; i < m; i++) {
          if (grid[i][j] >= val) {
            possible = true;
            if (st.find(i) != st.end()) {
              flag2 = true;
            }
            st.insert(i);
          }
        }
        flag1 &= possible;
      }
      return (flag1 && flag2);
    };
    long long s = 0, e = INF, res = s;
    while (s <= e) {
      long long mid = s + (e - s) / 2;
      if (isPossible(mid)) {
        res = mid;
        s = mid + 1;
      } else {
        e = mid - 1;
      }
    }
    cout << res << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  