#include <bits/stdc++.h>
using namespace std;
const int md = 1000000007;
long long int add(long long int a, long long int b) {
  return (a % md + b % md + md) % md;
}
long long int sub(long long int a, long long int b) {
  return (a % md + md - b % md) % md;
}
long long int mul(long long int a, long long int b) {
  return ((a % md) * (b % md) + md) % md;
}
const long long int INF = 1e18;
long long int dirA[] = {0, -1, 0, 1};
long long int dirB[] = {-1, 0, 1, 0};
long long int dirX[] = {0, -1, 0, 1, -1, -1, 1, 1};
long long int dirY[] = {-1, 0, 1, 0, -1, 1, -1, 1};
long long int power(long long int a, long long int b, long long int p) {
  if (a == 0) return 0;
  long long int res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
long long int min(long long int a, long long int b) { return (a < b) ? a : b; }
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 / gcd(a, b)) * b;
}
vector<bool> is_prime(100000001, true);
vector<long long int> primes;
void isPrime() {
  long long int n = 1000000;
  is_prime[0] = is_prime[1] = false;
  for (long long int i = 2; i <= n; i++) {
    if (is_prime[i] && i * i <= n) {
      for (long long int j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
bool sortbysec(const pair<long long int, long long int>& a,
               const pair<long long int, long long int>& b) {
  return (a.second < b.second);
}
vector<long long int> two;
void solve() {
  long long int m, n;
  cin >> m >> n;
  vector<vector<long long int> > joy(m, vector<long long int>(n, 0));
  for (long long int i = 0; i < m; i++) {
    for (long long int j = 0; j < n; j++) {
      cin >> joy[i][j];
    }
  }
  vector<long long int> ans(n);
  long long int final = 0;
  for (long long int i = 0; i < m; i++) {
    for (long long int j = 0; j < n; j++) {
      ans[j] = max(ans[j], joy[i][j]);
    }
  }
  if (m <= n - 1) {
    cout << *min_element((ans).begin(), (ans).end()) << "\n";
  } else {
    for (long long int i = 0; i < m; i++) {
      long long int max1 = 0, max2 = 0;
      long long int idx1 = 0, idx2 = 0;
      for (long long int j = 0; j < n; j++) {
        long long int x = joy[i][j];
        if (x > max1) {
          max2 = max1;
          max1 = x;
          idx2 = idx1;
          idx1 = j;
        } else {
          if (x > max2) {
            max2 = x;
            idx2 = j;
          }
        }
      }
      long long int a = ans[idx1], b = ans[idx2];
      ans[idx1] = max1;
      ans[idx2] = max2;
      final = max(final, *min_element((ans).begin(), (ans).end()));
      ans[idx1] = a, ans[idx2] = b;
    }
    cout << final << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  isPrime();
  long long int t = 1;
  cin >> t;
  long long int a = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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 md = 1000000007;
long long int add(long long int a, long long int b) {
  return (a % md + b % md + md) % md;
}
long long int sub(long long int a, long long int b) {
  return (a % md + md - b % md) % md;
}
long long int mul(long long int a, long long int b) {
  return ((a % md) * (b % md) + md) % md;
}
const long long int INF = 1e18;
long long int dirA[] = {0, -1, 0, 1};
long long int dirB[] = {-1, 0, 1, 0};
long long int dirX[] = {0, -1, 0, 1, -1, -1, 1, 1};
long long int dirY[] = {-1, 0, 1, 0, -1, 1, -1, 1};
long long int power(long long int a, long long int b, long long int p) {
  if (a == 0) return 0;
  long long int res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
long long int max(long long int a, long long int b) { return (a > b) ? a : b; }
long long int min(long long int a, long long int b) { return (a < b) ? a : b; }
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 / gcd(a, b)) * b;
}
vector<bool> is_prime(100000001, true);
vector<long long int> primes;
void isPrime() {
  long long int n = 1000000;
  is_prime[0] = is_prime[1] = false;
  for (long long int i = 2; i <= n; i++) {
    if (is_prime[i] && i * i <= n) {
      for (long long int j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
bool sortbysec(const pair<long long int, long long int>& a,
               const pair<long long int, long long int>& b) {
  return (a.second < b.second);
}
vector<long long int> two;
void solve() {
  long long int m, n;
  cin >> m >> n;
  vector<vector<long long int> > joy(m, vector<long long int>(n, 0));
  for (long long int i = 0; i < m; i++) {
    for (long long int j = 0; j < n; j++) {
      cin >> joy[i][j];
    }
  }
  vector<long long int> ans(n);
  long long int final = 0;
  for (long long int i = 0; i < m; i++) {
    for (long long int j = 0; j < n; j++) {
      ans[j] = max(ans[j], joy[i][j]);
    }
  }
  if (m <= n - 1) {
    cout << *min_element((ans).begin(), (ans).end()) << "\n";
  } else {
    for (long long int i = 0; i < m; i++) {
      long long int max1 = 0, max2 = 0;
      long long int idx1 = 0, idx2 = 0;
      for (long long int j = 0; j < n; j++) {
        long long int x = joy[i][j];
        if (x > max1) {
          max2 = max1;
          max1 = x;
          idx2 = idx1;
          idx1 = j;
        } else {
          if (x > max2) {
            max2 = x;
            idx2 = j;
          }
        }
      }
      long long int a = ans[idx1], b = ans[idx2];
      ans[idx1] = max1;
      ans[idx2] = max2;
      final = max(final, *min_element((ans).begin(), (ans).end()));
      ans[idx1] = a, ans[idx2] = b;
    }
    cout << final << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  isPrime();
  long long int t = 1;
  cin >> t;
  long long int a = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```