#include <bits/stdc++.h>
using namespace std;
long long t = 1, n, m, k, q, l, r, a[100009], b, c, u, wtf[100009], x, y, z,
          mod = 1e9 + 7;
string s;
vector<long long> v[100009];
struct st {
  long long x, y;
};
bool cmp(const st &a, const st &b) { return a.x < b.x; }
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0')
    s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long qp(long long x, long long y) {
  long long a = 1, b = x;
  while (y) {
    if (y & 1) a = a * b % mod;
    b = b * b % mod, y >>= 1;
  }
  return a;
}
bool judge(long long x) {
  fill(wtf + 1, wtf + m + 1, 0);
  bool qwq = 0;
  for (long long i = 1; i <= n; i += 1) {
    bool fl = 0;
    for (long long j = 1; j <= m; j += 1) {
      if (v[j][i] >= x && wtf[j]) qwq = 1;
      if (v[j][i] >= x) wtf[j] = 1, fl = 1;
    }
    if (!fl) return 0;
  }
  return qwq;
}
int main() {
  t = read();
  while (t--) {
    m = read(), n = read();
    fill(a + 1, a + n + 1, 0);
    for (long long i = 1; i <= m; i += 1) v[i].resize(n + 2);
    for (long long i = 1; i <= m; i += 1)
      for (long long j = 1; j <= n; j += 1) v[i][j] = read();
    if (n > m) {
      for (long long i = 1; i <= m; i += 1)
        for (long long j = 1; j <= n; j += 1) a[j] = max(a[j], v[i][j]);
      printf("%lld\n", *min_element(a + 1, a + n + 1));
    } else {
      long long l = 1, r = 1e9, ans = 0;
      while (l <= r) {
        long long mid = (l + r) >> 1;
        if (judge(mid))
          ans = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      printf("%lld\n", ans);
    }
  }
  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;
long long t = 1, n, m, k, q, l, r, a[100009], b, c, u, wtf[100009], x, y, z,
          mod = 1e9 + 7;
string s;
vector<long long> v[100009];
struct st {
  long long x, y;
};
bool cmp(const st &a, const st &b) { return a.x < b.x; }
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0')
    s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  return s * w;
}
long long qp(long long x, long long y) {
  long long a = 1, b = x;
  while (y) {
    if (y & 1) a = a * b % mod;
    b = b * b % mod, y >>= 1;
  }
  return a;
}
bool judge(long long x) {
  fill(wtf + 1, wtf + m + 1, 0);
  bool qwq = 0;
  for (long long i = 1; i <= n; i += 1) {
    bool fl = 0;
    for (long long j = 1; j <= m; j += 1) {
      if (v[j][i] >= x && wtf[j]) qwq = 1;
      if (v[j][i] >= x) wtf[j] = 1, fl = 1;
    }
    if (!fl) return 0;
  }
  return qwq;
}
int main() {
  t = read();
  while (t--) {
    m = read(), n = read();
    fill(a + 1, a + n + 1, 0);
    for (long long i = 1; i <= m; i += 1) v[i].resize(n + 2);
    for (long long i = 1; i <= m; i += 1)
      for (long long j = 1; j <= n; j += 1) v[i][j] = read();
    if (n > m) {
      for (long long i = 1; i <= m; i += 1)
        for (long long j = 1; j <= n; j += 1) a[j] = max(a[j], v[i][j]);
      printf("%lld\n", *min_element(a + 1, a + n + 1));
    } else {
      long long l = 1, r = 1e9, ans = 0;
      while (l <= r) {
        long long mid = (l + r) >> 1;
        if (judge(mid))
          ans = mid, l = mid + 1;
        else
          r = mid - 1;
      }
      printf("%lld\n", ans);
    }
  }
  return 0;
}

```