#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
void fileio(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const long long INF = 4e18;
inline long long read() {
  long long x = 0;
  bool flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return (flag ? x : ~(x - 1));
}
long long t, n, m, vis[100000], mst[100000], num[100000];
vector<long long> p[100000], ok[100000];
bool check(long long x) {
  fill(vis, vis + m, 0);
  fill(mst, mst + n, 0);
  fill(num, num + n, 0);
  for (long long i = 0; i < m; i++) ok[i].clear();
  for (long long i = 0; i < m; i++)
    for (long long l = 0; l < n; l++) ok[i].push_back(0);
  for (long long i = 0; i < n; i++)
    for (long long l = 0; l < m; l++)
      if (p[i][l] >= x) {
        ok[l][i] = 1;
        vis[l]++;
        num[i]++;
      }
  for (long long i = 0; i < m; i++)
    if (!vis[i]) return 0;
  for (long long i = 0; i < m; i++)
    for (long long l = 0; l < n; l++)
      if (vis[i] == 1 && ok[i][l]) mst[l] = 1;
  long long res = 0;
  for (long long i = 0; i < n; i++) res += mst[i];
  if (res >= m) return 0;
  for (long long i = 0; i < n; i++)
    if (num[i] > 1) return 1;
  return 0;
}
void solve() {
  n = read();
  m = read();
  for (long long i = 0; i < n; i++) {
    p[i].clear();
    for (long long l = 1; l <= m; l++) p[i].push_back(read());
  }
  long long l = 1, r = 1000000000;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  cout << l - 1 << endl;
}
signed main() {
  t = read();
  while (t--) solve();
  return 0;
}

### 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>
#pragma GCC optimize("Ofast")
using namespace std;
void fileio(const string &s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const long long INF = 4e18;
inline long long read() {
  long long x = 0;
  bool flag = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') flag = 0;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return (flag ? x : ~(x - 1));
}
long long t, n, m, vis[100000], mst[100000], num[100000];
vector<long long> p[100000], ok[100000];
bool check(long long x) {
  fill(vis, vis + m, 0);
  fill(mst, mst + n, 0);
  fill(num, num + n, 0);
  for (long long i = 0; i < m; i++) ok[i].clear();
  for (long long i = 0; i < m; i++)
    for (long long l = 0; l < n; l++) ok[i].push_back(0);
  for (long long i = 0; i < n; i++)
    for (long long l = 0; l < m; l++)
      if (p[i][l] >= x) {
        ok[l][i] = 1;
        vis[l]++;
        num[i]++;
      }
  for (long long i = 0; i < m; i++)
    if (!vis[i]) return 0;
  for (long long i = 0; i < m; i++)
    for (long long l = 0; l < n; l++)
      if (vis[i] == 1 && ok[i][l]) mst[l] = 1;
  long long res = 0;
  for (long long i = 0; i < n; i++) res += mst[i];
  if (res >= m) return 0;
  for (long long i = 0; i < n; i++)
    if (num[i] > 1) return 1;
  return 0;
}
void solve() {
  n = read();
  m = read();
  for (long long i = 0; i < n; i++) {
    p[i].clear();
    for (long long l = 1; l <= m; l++) p[i].push_back(read());
  }
  long long l = 1, r = 1000000000;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (check(mid))
      l = mid + 1;
    else
      r = mid - 1;
  }
  cout << l - 1 << endl;
}
signed main() {
  t = read();
  while (t--) solve();
  return 0;
}

```