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 mod = 1e9 + 7;
  const int N = 1e6 + 4;
  void _print(long long t) { cerr << t; }
  void _print(int t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  long long po(long long x, long long n) {
    long long ans = 1;
    while (n) {
      if (n % 2 == 1) ans = ans * x;
      x = x * x;
      n = n / 2;
    }
    return ans;
  }
  long long lo(long long k) {
    long long ans = 0;
    while (k > 1) {
      k = k / 2;
      ans++;
    }
    return ans;
  }
  void JAI_SHREE_RAM() {
    long long n, m;
    cin >> n >> m;
    vector<vector<long long>> a(n, vector<long long>(m, 0));
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        cin >> a[i][j];
      }
    long long ch = m - 1;
    if (ch >= n) {
      long long ans = LLONG_MAX;
      for (int i = 0; i < m; i++) {
        long long mx = -100000000000000ll;
        for (int j = 0; j < n; j++) {
          mx = max(a[j][i], mx);
        }
        ans = min(ans, mx);
      }
      cout << ans << "\n";
      return;
    }
    vector<pair<long long, long long>> p;
    long long row = 0;
    long long val = -100000000000000ll;
    for (int i = 0; i < n; i++) {
      vector<pair<long long, long long>> v;
      for (int j = 0; j < m; j++) {
        v.push_back({a[i][j], j});
      }
      sort(v.begin(), v.end(), greater<pair<long long, long long>>());
      if (v[1].first > val) {
        val = v[1].first;
        row = i;
      }
    }
    set<long long> col;
    vector<pair<long long, long long>> temp;
    for (int i = 0; i < m; i++) temp.push_back({a[row][i], i});
    sort(temp.begin(), temp.end(), greater<pair<long long, long long>>());
    col.insert(temp[0].second);
    col.insert(temp[1].second);
    ;
    long long ans = min(temp[1].first, temp[0].first);
    ;
    ;
    for (int i = 0; i < m; i++) {
      if (col.find(i) != col.end()) continue;
      long long mx = -100000000000000ll;
      for (int j = 0; j < n; j++) {
        mx = max(a[j][i], mx);
      }
      ans = min(ans, mx);
    }
    cout << ans << "\n";
    return;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long tt = 1;
    cin >> tt;
    for (int i = 1; i < tt + 1; i++) {
      JAI_SHREE_RAM();
    }
  }
  