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 warning disable format
  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 gcd(long long a, long long b) {
    if (b > a) {
      return gcd(b, a);
    }
    if (b == 0) {
      return a;
    }
    return gcd(b, a % b);
  }
  long long powm(long long a, long long b, long long mod) {
    long long res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b = b >> 1;
    }
    return res;
  }
  void extendgcd(long long a, long long b, long long* v) {
    if (b == 0) {
      v[0] = 1;
      v[1] = 0;
      v[2] = a;
      return;
    }
    extendgcd(b, a % b, v);
    long long x = v[1];
    v[1] = v[0] - v[1] * (a / b);
    v[0] = x;
    return;
  }
  long long mminv(long long a, long long b) {
    long long arr[3];
    extendgcd(a, b, arr);
    return arr[0];
  }
  long long mminvprime(long long a, long long b) { return powm(a, b - 2, b); }
  vector<long long> sieve(int n) {
    int* arr = new int[n + 1]();
    vector<long long> vect;
    for (int i = 2; i <= n; i++)
      if (arr[i] == 0) {
        vect.push_back(i);
        for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
      }
    return vect;
  }
  vector<string> split(string s, char delimeter) {
    vector<string> tokens;
    stringstream check1(s);
    string intermediate;
    while (getline(check1, intermediate, delimeter)) {
      tokens.push_back(intermediate);
    }
    return tokens;
  }
  void toLower(string& s) { transform(s.begin(), s.end(), s.begin(), ::tolower); }
  void toUpper(string& s) { transform(s.begin(), s.end(), s.begin(), ::toupper); }
  long long addm(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a + b) % m) + m) % m;
  }
  long long mulm(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a * b) % m) + m) % m;
  }
  long long subm(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (((a - b) % m) + m) % m;
  }
  long long divm(long long a, long long b, long long m) {
    a = a % m;
    b = b % m;
    return (mulm(a, mminvprime(b, m), m) + m) % m;
  }
  const long long SIZE = 2;
  long long fact[SIZE], ifact[SIZE];
  void gen_factorial(long long n, long long mod) {
    fact[0] = fact[1] = ifact[0] = ifact[1] = 1;
    for (long long i = 2; i <= n; i++) {
      fact[i] = (i * fact[i - 1]) % mod;
    }
    ifact[n] = mminvprime(fact[n], mod);
    for (long long i = n - 1; i >= 2; i--) {
      ifact[i] = ((i + 1) * ifact[i + 1]) % mod;
    }
  }
  long long choose(long long n, long long r, long long m) {
    long long val1 = fact[n];
    long long val2 = ifact[n - r];
    long long val3 = ifact[r];
    return (((val1 * val2) % m) * val3) % m;
  }
  long long i, j, k;
  const int MOD = 1000000007;
  const int MOD1 = 998244353;
  #pragma warning restore format
  bool isPossible(long long joy, vector<vector<long long> >& p, int n, int m) {
    set<int> visited;
    if ((m <= n - 1)) {
      for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
          if (p[i][j] >= joy) visited.insert(j);
        }
      }
      return (visited.size() == n);
    }
    int index = -1;
    for (int i = 0; i < m; i++) {
      int count = 0;
      for (int j = 0; j < n; j++) {
        if (p[i][j] >= joy) count++;
      }
      if (count >= 2) {
        index = i;
        break;
      }
    }
    if (index == -1) return false;
    for (int j = 0; j < n; j++) {
      if (p[index][j] >= joy) visited.insert(j);
    }
    for (int i = 0; i < m; i++) {
      bool flag = false;
      if (i == index) continue;
      for (int j = 0; j < n; j++) {
        if ((p[i][j] >= joy) && (visited.find(j) == visited.end())) {
          flag = true;
          break;
        }
      }
      if (!flag) continue;
      for (int j = 0; j < n; j++) {
        if (p[i][j] >= joy) visited.insert(j);
      }
    }
    return (visited.size() == n);
  }
  void solve() {
    int n, m;
    cin >> m >> n;
    vector<vector<long long> > p(m, vector<long long>(n));
    long long low = 0, high, maxE = 0;
    for (int i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        cin >> p[i][j];
        maxE = max(maxE, p[i][j]);
      }
    };
    high = maxE;
    long long ans = 0;
    while (low <= high) {
      ;
      int mid = (low + high) / 2;
      ;
      if (isPossible(mid, p, n, m)) {
        ans = mid;
        low = mid + 1;
      } else
        high = mid - 1;
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int T = 1;
    cin >> T;
    while (T--) {
      solve();
    }
    return 0;
  }
  