#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return (gcd(y % x, x));
}
string Tolower(string s) {
  transform(s.begin(), s.end(), s.begin(), ::tolower);
  return s;
}
string Toupper(string s) {
  transform(s.begin(), s.end(), s.begin(), ::toupper);
  return s;
}
void solve() {
  long long t;
  cin >> t;
  while (t--) {
    long long m, n, ans = -1, l, r, mid;
    cin >> m >> n;
    vector<vector<long long>> a(m);
    set<long long> s;
    vector<long long> b;
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        long long x;
        cin >> x;
        a[i].push_back(x);
        s.insert(x);
      }
    }
    for (auto x : s) b.push_back(x);
    l = 0, r = b.size() - 1;
    while (l <= r) {
      mid = l + (r - l) / 2;
      bool f = true;
      map<long long, long long> make_pair;
      set<long long> sex;
      for (long long i = 0; i < m; i++) {
        for (long long j = 0; j < n; j++) {
          if (a[i][j] >= b[mid]) {
            make_pair[i]++;
            sex.insert(j);
          }
        }
      }
      vector<long long> temp;
      for (auto x : make_pair) temp.push_back(x.second);
      sort((temp).begin(), (temp).end(), greater<int>());
      long long sum = 0;
      long long sz = temp.size();
      for (long long i = 0; i < min(n - 1, sz); i++) {
        sum += temp[i];
      }
      if (sum >= n && (long long)sex.size() == n)
        f = true;
      else
        f = false;
      if (f) {
        ans = b[mid];
        l = mid + 1;
      } else
        r = mid - 1;
    }
    if (n - 1 >= m) {
      ans = 1e9;
      for (long long i = 0; i < n; i++) {
        long long mx = -1;
        for (long long j = 0; j < m; j++) {
          mx = max(mx, a[j][i]);
        }
        ans = min(ans, mx);
      }
    }
    cout << ans << "\n";
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return (gcd(y % x, x));
}
string Tolower(string s) {
  transform(s.begin(), s.end(), s.begin(), ::tolower);
  return s;
}
string Toupper(string s) {
  transform(s.begin(), s.end(), s.begin(), ::toupper);
  return s;
}
void solve() {
  long long t;
  cin >> t;
  while (t--) {
    long long m, n, ans = -1, l, r, mid;
    cin >> m >> n;
    vector<vector<long long>> a(m);
    set<long long> s;
    vector<long long> b;
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        long long x;
        cin >> x;
        a[i].push_back(x);
        s.insert(x);
      }
    }
    for (auto x : s) b.push_back(x);
    l = 0, r = b.size() - 1;
    while (l <= r) {
      mid = l + (r - l) / 2;
      bool f = true;
      map<long long, long long> make_pair;
      set<long long> sex;
      for (long long i = 0; i < m; i++) {
        for (long long j = 0; j < n; j++) {
          if (a[i][j] >= b[mid]) {
            make_pair[i]++;
            sex.insert(j);
          }
        }
      }
      vector<long long> temp;
      for (auto x : make_pair) temp.push_back(x.second);
      sort((temp).begin(), (temp).end(), greater<int>());
      long long sum = 0;
      long long sz = temp.size();
      for (long long i = 0; i < min(n - 1, sz); i++) {
        sum += temp[i];
      }
      if (sum >= n && (long long)sex.size() == n)
        f = true;
      else
        f = false;
      if (f) {
        ans = b[mid];
        l = mid + 1;
      } else
        r = mid - 1;
    }
    if (n - 1 >= m) {
      ans = 1e9;
      for (long long i = 0; i < n; i++) {
        long long mx = -1;
        for (long long j = 0; j < m; j++) {
          mx = max(mx, a[j][i]);
        }
        ans = min(ans, mx);
      }
    }
    cout << ans << "\n";
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```