#include <bits/stdc++.h>
using namespace std;
const int mod = 1e9 + 7;
const int N = 2e5 + 17;
void solve() {
  long long n, m;
  cin >> m >> n;
  vector<vector<long long>> p(n, vector<long long>(m)),
      a(n, vector<long long>(m)), need(m);
  vector<long long> good(m);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      cin >> p[j][i];
    }
  }
  long long l = -1, r = 1e18;
  while (r - l > 1) {
    long long mid = (r + l) / 2;
    bool ok = 1;
    vector<long long> cnt(m, 0);
    for (int i = 0; i < n; i++) {
      bool flag = 0;
      for (int j = 0; j < m; j++) {
        if (p[i][j] >= mid) {
          cnt[j]++;
          flag = 1;
        }
      }
      ok &= flag;
    }
    long long mx = *max_element((cnt).begin(), (cnt).end());
    if (mx > 1 && ok) {
      l = mid;
    } else {
      r = mid;
    }
  }
  cout << l << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Your challenge is to write 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;
const int mod = 1e9 + 7;
const int N = 2e5 + 17;
void solve() {
  long long n, m;
  cin >> m >> n;
  vector<vector<long long>> p(n, vector<long long>(m)),
      a(n, vector<long long>(m)), need(m);
  vector<long long> good(m);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < n; j++) {
      cin >> p[j][i];
    }
  }
  long long l = -1, r = 1e18;
  while (r - l > 1) {
    long long mid = (r + l) / 2;
    bool ok = 1;
    vector<long long> cnt(m, 0);
    for (int i = 0; i < n; i++) {
      bool flag = 0;
      for (int j = 0; j < m; j++) {
        if (p[i][j] >= mid) {
          cnt[j]++;
          flag = 1;
        }
      }
      ok &= flag;
    }
    long long mx = *max_element((cnt).begin(), (cnt).end());
    if (mx > 1 && ok) {
      l = mid;
    } else {
      r = mid;
    }
  }
  cout << l << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```