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 ull = unsigned long long;
  using ll = long long;
  using u64 = uint64_t;
  using u32 = uint32_t;
  using i64 = int64_t;
  using i32 = int32_t;
  using namespace std;
  template <typename T>
  void print_vector(ostream &os, const vector<T> &v, const string &sep = " ",
                    const string &begin = "", const string &end = "\n") {
    os << begin;
    for (size_t i = 0, n{v.size()}; i < n; ++i) {
      if (i < n - 1) {
        os << v[i] << sep;
      } else {
        os << v[i];
      }
    }
    os << end;
  }
  template <typename T>
  ostream &operator<<(ostream &os, const vector<T> &v) {
    print_vector(os, v, ", ", "[", "]");
    return os;
  }
  template <typename _Key, typename _Compare = std::less<_Key>,
            typename _Alloc = std::allocator<_Key>>
  ostream &operator<<(ostream &os, const set<_Key, _Compare, _Alloc> &s) {
    os << '{';
    if (s.size()) {
      auto pend = begin(s);
      advance(pend, s.size() - 1);
      for (auto it = begin(s); it != pend; ++it) {
        os << *it << ", ";
      }
      os << *s.rbegin();
    }
    return os << '}';
  }
  template <typename _Value, typename _Hash = hash<_Value>,
            typename _Pred = equal_to<_Value>,
            typename _Alloc = allocator<_Value>>
  ostream &operator<<(ostream &os,
                      const unordered_set<_Value, _Hash, _Pred, _Alloc> &s) {
    os << '{';
    if (s.size()) {
      auto pend = begin(s);
      advance(pend, s.size() - 1);
      for (auto it = begin(s); it != pend; ++it) {
        os << *it;
        if (it != pend) {
          os << ", ";
        }
      }
    }
    return os << '}';
  }
  template <typename T, typename U>
  ostream &operator<<(ostream &os, const pair<T, U> &p) {
    return os << '(' << p.first << ',' << p.second << ')';
  }
  template <typename K, typename V>
  ostream &operator<<(ostream &os, const map<K, V> &m) {
    os << '{';
    if (m.size()) {
      auto pend = begin(m);
      advance(pend, m.size() - 1);
      for (auto it = begin(m); it != pend; ++it) {
        os << *it << ", ";
      }
      os << m.back();
    }
    return os << '}';
  }
  template <typename K, typename V>
  ostream &operator<<(ostream &os, const unordered_map<K, V> &m) {
    os << '{';
    if (m.size()) {
      auto pend = begin(m);
      advance(pend, m.size() - 1);
      for (auto it = begin(m); it != pend; ++it) {
        os << *it << ", ";
      }
      os << m.back();
    }
    return os << '}';
  }
  template <class T>
  inline void hash_combine(std::size_t &seed, T const &v) {
    seed ^= std::hash<T>()(v) + 0x9e3779b9 + (seed << 6) + (seed >> 2);
  }
  template <class Tuple, size_t Index = std::tuple_size<Tuple>::value - 1>
  struct HashValueImpl {
    static void apply(size_t &seed, Tuple const &tuple) {
      HashValueImpl<Tuple, Index - 1>::apply(seed, tuple);
      hash_combine(seed, std::get<Index>(tuple));
    }
  };
  template <class Tuple>
  struct HashValueImpl<Tuple, 0> {
    static void apply(size_t &seed, Tuple const &tuple) {
      hash_combine(seed, std::get<0>(tuple));
    }
  };
  template <typename... TT>
  struct hash<std::tuple<TT...>> {
    size_t operator()(std::tuple<TT...> const &tt) const {
      size_t seed = 0;
      HashValueImpl<std::tuple<TT...>>::apply(seed, tt);
      return seed;
    }
  };
  struct hash_pair {
    template <class T1, class T2>
    size_t operator()(const pair<T1, T2> &p) const {
      auto hash1 = hash<T1>{}(p.first);
      auto hash2 = hash<T2>{}(p.second);
      return hash1 ^ hash2;
    }
  };
  void solve() {
    int m = 1, n = 1;
    cin >> m >> n;
    vector<vector<ll>> v(m, vector<ll>(n));
    for (auto &r : v) {
      for (auto &x : r) {
        cin >> x;
      }
    }
    auto test = [&](ll threshold) {
      vector<bool> possible(n, false);
      bool has_pair = false;
      for (int i = 0; i < m; ++i) {
        int satisfied_cols = 0;
        for (int j = 0; j < n; ++j) {
          possible[j] = possible[j] or v[i][j] >= threshold;
          satisfied_cols += v[i][j] >= threshold;
        }
        has_pair = has_pair or satisfied_cols >= 2;
      }
      return has_pair and all_of((possible).begin(), (possible).end(),
                                 [](bool x) { return x; });
    };
    ll lo = 0, hi = ll(1e9) + 1;
    while (lo < hi) {
      ll mid = lo + (hi - lo + 1) / 2;
      if (test(mid)) {
        lo = mid;
      } else {
        hi = mid - 1;
      }
    }
    cout << lo << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cerr.tie(NULL);
    int t = 1;
    cin >> t;
    for (int tt = 1; tt <= t; ++tt) {
      solve();
    }
    return 0;
  }
  