#include <bits/stdc++.h>
using namespace std;
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool comp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int main() {
  fast();
  int t, m, n, i, j, res1, res2, mn, mx, ans;
  string s;
  cin >> t;
  while (t--) {
    cin >> m >> n;
    mx = 1000000007;
    vector<vector<int>> v(m, vector<int>(n));
    vector<vector<pair<int, int>>> vv(n);
    set<int> st;
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) cin >> v[i][j];
    }
    ans = 0;
    for (i = 0; i < m; i++) {
      vector<int> vvv = v[i];
      sort(vvv.begin(), vvv.end(), greater<int>());
      ans = max(ans, vvv[1]);
    }
    for (i = 0; i < n; i++) {
      mx = 0;
      for (j = 0; j < m; j++) {
        mx = max(mx, v[j][i]);
      }
      ans = min(mx, ans);
    }
    cout << ans << '\n';
  }
}

### Prompt

Please create a solution in cpp 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;
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool comp(pair<int, int> a, pair<int, int> b) {
  if (a.first == b.first) return a.second < b.second;
  return a.first > b.first;
}
int main() {
  fast();
  int t, m, n, i, j, res1, res2, mn, mx, ans;
  string s;
  cin >> t;
  while (t--) {
    cin >> m >> n;
    mx = 1000000007;
    vector<vector<int>> v(m, vector<int>(n));
    vector<vector<pair<int, int>>> vv(n);
    set<int> st;
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) cin >> v[i][j];
    }
    ans = 0;
    for (i = 0; i < m; i++) {
      vector<int> vvv = v[i];
      sort(vvv.begin(), vvv.end(), greater<int>());
      ans = max(ans, vvv[1]);
    }
    for (i = 0; i < n; i++) {
      mx = 0;
      for (j = 0; j < m; j++) {
        mx = max(mx, v[j][i]);
      }
      ans = min(mx, ans);
    }
    cout << ans << '\n';
  }
}

```