#include <bits/stdc++.h>
using namespace std;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  long long int p = sqrt(n);
  for (int i = 5; i <= p; i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
string d2b(int n) {
  string s;
  for (int i = 31; i >= 0; i--) {
    int k = n >> i;
    if (k & 1)
      s += '1';
    else
      s += "0";
  }
  return s;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int b2d(string n) {
  string num = n;
  long long int dec_value = 0;
  long long int base = 1;
  long long int len = num.length();
  for (int i = len - 1; i >= 0; i--) {
    if (num[i] == '1') dec_value += base;
    base = base * 2;
  }
  return dec_value;
}
bool check(long long int x, vector<vector<long long int>> &v) {
  long long int m = v.size();
  long long int n = v[0].size();
  unordered_map<long long int, long long int> mp;
  for (long long j = 0; j < n; j++) {
    bool f = 0;
    for (long long i = 0; i < m; i++) {
      if (v[i][j] >= x) {
        f = 1;
        break;
      }
    }
    if (f == 0) return 0;
  }
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (v[i][j] >= x) {
        if (mp[i] == 1) {
          return 1;
        }
        mp[i] = 1;
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test_case = 1;
  cin >> test_case;
  while (test_case--) {
    long long int n, m;
    cin >> m >> n;
    vector<vector<long long int>> a(m, vector<long long int>(n));
    long long int l = INT_MAX, h = INT_MIN;
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        cin >> a[i][j];
        l = min(l, a[i][j]);
        h = max(h, a[i][j]);
      }
    }
    long long int mid;
    long long int ans = 0;
    while (l <= h) {
      mid = (l + h) / 2;
      if (check(mid, a)) {
        ans = mid;
        l = mid + 1;
      } else
        h = mid - 1;
    }
    cout << ans << endl;
  }
}

### Prompt

In Cpp, your task is to solve 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;
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  long long int p = sqrt(n);
  for (int i = 5; i <= p; i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
string d2b(int n) {
  string s;
  for (int i = 31; i >= 0; i--) {
    int k = n >> i;
    if (k & 1)
      s += '1';
    else
      s += "0";
  }
  return s;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int b2d(string n) {
  string num = n;
  long long int dec_value = 0;
  long long int base = 1;
  long long int len = num.length();
  for (int i = len - 1; i >= 0; i--) {
    if (num[i] == '1') dec_value += base;
    base = base * 2;
  }
  return dec_value;
}
bool check(long long int x, vector<vector<long long int>> &v) {
  long long int m = v.size();
  long long int n = v[0].size();
  unordered_map<long long int, long long int> mp;
  for (long long j = 0; j < n; j++) {
    bool f = 0;
    for (long long i = 0; i < m; i++) {
      if (v[i][j] >= x) {
        f = 1;
        break;
      }
    }
    if (f == 0) return 0;
  }
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (v[i][j] >= x) {
        if (mp[i] == 1) {
          return 1;
        }
        mp[i] = 1;
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test_case = 1;
  cin >> test_case;
  while (test_case--) {
    long long int n, m;
    cin >> m >> n;
    vector<vector<long long int>> a(m, vector<long long int>(n));
    long long int l = INT_MAX, h = INT_MIN;
    for (long long i = 0; i < m; i++) {
      for (long long j = 0; j < n; j++) {
        cin >> a[i][j];
        l = min(l, a[i][j]);
        h = max(h, a[i][j]);
      }
    }
    long long int mid;
    long long int ans = 0;
    while (l <= h) {
      mid = (l + h) / 2;
      if (check(mid, a)) {
        ans = mid;
        l = mid + 1;
      } else
        h = mid - 1;
    }
    cout << ans << endl;
  }
}

```