#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
void pr() { cout << endl; }
template <typename T1, typename... T2>
void pr(const T1& arg, const T2&... args) {
  cout << arg << " ";
  pr(args...);
}
vector<vector<int> > v;
int m, n;
bool pd(int x) {
  int book[m + 5];
  memset(book, 0, sizeof(book));
  for (int j = 1; j <= n; ++j) {
    bool flg = 0;
    for (int i = 1; i <= m; ++i) {
      if (v[i][j] >= x) {
        book[i]++;
        flg = 1;
      }
    }
    if (!flg) return 0;
  }
  for (int i = 1; i <= m; ++i) {
    if (book[i] >= 2) return 1;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    cin >> m >> n;
    v.clear();
    v.resize(m + 2, vector<int>(n + 2));
    int l = 1, r = 0;
    for (int i = 1; i <= m; ++i) {
      for (int j = 1; j <= n; ++j) {
        cin >> v[i][j];
        r = max(r, v[i][j]);
      }
    }
    while (r - l > 1) {
      int m = (l + r) >> 1;
      if (pd(m))
        l = m;
      else
        r = m - 1;
    }
    if (pd(r))
      cout << r << '\n';
    else
      cout << l << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 inf = 0x3f3f3f3f;
void pr() { cout << endl; }
template <typename T1, typename... T2>
void pr(const T1& arg, const T2&... args) {
  cout << arg << " ";
  pr(args...);
}
vector<vector<int> > v;
int m, n;
bool pd(int x) {
  int book[m + 5];
  memset(book, 0, sizeof(book));
  for (int j = 1; j <= n; ++j) {
    bool flg = 0;
    for (int i = 1; i <= m; ++i) {
      if (v[i][j] >= x) {
        book[i]++;
        flg = 1;
      }
    }
    if (!flg) return 0;
  }
  for (int i = 1; i <= m; ++i) {
    if (book[i] >= 2) return 1;
  }
  return 0;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    cin >> m >> n;
    v.clear();
    v.resize(m + 2, vector<int>(n + 2));
    int l = 1, r = 0;
    for (int i = 1; i <= m; ++i) {
      for (int j = 1; j <= n; ++j) {
        cin >> v[i][j];
        r = max(r, v[i][j]);
      }
    }
    while (r - l > 1) {
      int m = (l + r) >> 1;
      if (pd(m))
        l = m;
      else
        r = m - 1;
    }
    if (pd(r))
      cout << r << '\n';
    else
      cout << l << '\n';
  }
  return 0;
}

```