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 T, size_t N>
  long long SIZE(const T (&t)[N]) {
    return N;
  }
  template <typename T>
  long long SIZE(const T &t) {
    return t.size();
  }
  string to_string(const string s, long long x1 = 0, long long x2 = 1e9) {
    return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
  }
  string to_string(const char *s) { return to_string((string)s); }
  string to_string(const bool b) { return (b ? "true" : "false"); }
  string to_string(const char c) { return string({c}); }
  template <size_t N>
  string to_string(const bitset<N> &b, long long x1 = 0, long long x2 = 1e9) {
    string t = "";
    for (long long __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
         __iii__ <= __jjj__; ++__iii__) {
      t += b[__iii__] + '0';
    }
    return '"' + t + '"';
  }
  template <typename A, typename... C>
  string to_string(const A(&v), long long x1 = 0, long long x2 = 1e9,
                   C... coords);
  long long l_v_l_v_l = 0, t_a_b_s = 0;
  template <typename A, typename B>
  string to_string(const pair<A, B> &p) {
    l_v_l_v_l++;
    string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
    l_v_l_v_l--;
    return res;
  }
  template <typename A, typename... C>
  string to_string(const A(&v), long long x1, long long x2, C... coords) {
    long long rnk = rank<A>::value;
    string tab(t_a_b_s, ' ');
    string res = "";
    bool first = true;
    if (l_v_l_v_l == 0) res += "\n";
    res += tab + "[";
    x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
    auto l = begin(v);
    advance(l, x1);
    auto r = l;
    advance(r, (x2 - x1) + (x2 < SIZE(v)));
    for (auto e = l; e != r; e = next(e)) {
      if (!first) {
        res += ", ";
      }
      first = false;
      l_v_l_v_l++;
      if (e != l) {
        if (rnk > 1) {
          res += "\n";
          t_a_b_s = l_v_l_v_l;
        };
      } else {
        t_a_b_s = 0;
      }
      res += to_string(*e, coords...);
      l_v_l_v_l--;
    }
    res += "]";
    if (l_v_l_v_l == 0) res += "\n";
    return res;
  }
  void dbgm() { ; }
  template <typename Heads, typename... Tails>
  void dbgm(Heads H, Tails... T) {
    cerr << to_string(H) << " | ";
    dbgm(T...);
  }
  void FIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
  }
  const long long N = 2e5 + 6;
  long long n, m;
  vector<vector<long long> > a;
  bool check(long long x) {
    vector<long long> v(m, 0);
    bool ok = 0;
    for (long long i = 0; i < n; i++) {
      long long c = 0;
      for (long long j = 0; j < m; j++) {
        if (a[i][j] >= x) {
          v[j] = 1;
          c++;
        }
      }
      if (c >= 2) ok = 1;
    }
    if (m > 1 and ok == 0) return 0;
    for (long long i = 0; i < m; i++)
      if (!v[i]) return 0;
    return 1;
  }
  int32_t main() {
    FIO();
    long long tc;
    cin >> tc;
    while (tc--) {
      cin >> n >> m;
      a.assign(n, vector<long long>(m));
      for (long long i = 0; i < n; i++)
        for (long long j = 0; j < m; j++) cin >> a[i][j];
      cerr << "["
           << "a"
           << "]: ";
      cerr << to_string(a) << endl;
      long long low = 1, high = 2e18, ans = -1;
      while (low <= high) {
        long long mid = (low + high) / 2;
        if (check(mid)) {
          ans = max(ans, mid);
          low = mid + 1;
        } else
          high = mid - 1;
      }
      cout << ans << '\n';
    }
    return 0;
  }
  