#include <bits/stdc++.h>
using namespace std;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
vector<int> g[N];
int a[N];
void solve() {
  int i, j, n, m;
  cin >> m >> n;
  vector<vector<int> > vec(m, vector<int>(n));
  vector<int> mx(n, INT_MIN);
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      cin >> vec[i][j];
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      mx[i] = max(mx[i], vec[j][i]);
    }
  }
  for (i = 0; i < m; i++) {
    sort(vec[i].begin(), vec[i].end());
  }
  int maxi = INT_MIN;
  for (i = 0; i < m; i++) {
    maxi = max(maxi, vec[i][n - 2]);
  }
  int ans = INT_MAX;
  for (i = 0; i < n; i++) {
    ans = min(ans, mx[i]);
  }
  ans = min(ans, maxi);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
void ipgraph(int n, int m) {
  int i, u, v;
  while (m--) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
void dfs(int u, int par) {
  for (int v : g[u]) {
    if (v == par) continue;
    dfs(v, u);
  }
}

### Prompt

Create a solution in Cpp for 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;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
mt19937_64 rang(
    chrono::high_resolution_clock::now().time_since_epoch().count());
int rng(int lim) {
  uniform_int_distribution<int> uid(0, lim - 1);
  return uid(rang);
}
int mpow(int base, int exp);
void ipgraph(int n, int m);
void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
vector<int> g[N];
int a[N];
void solve() {
  int i, j, n, m;
  cin >> m >> n;
  vector<vector<int> > vec(m, vector<int>(n));
  vector<int> mx(n, INT_MIN);
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      cin >> vec[i][j];
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      mx[i] = max(mx[i], vec[j][i]);
    }
  }
  for (i = 0; i < m; i++) {
    sort(vec[i].begin(), vec[i].end());
  }
  int maxi = INT_MIN;
  for (i = 0; i < m; i++) {
    maxi = max(maxi, vec[i][n - 2]);
  }
  int ans = INT_MAX;
  for (i = 0; i < n; i++) {
    ans = min(ans, mx[i]);
  }
  ans = min(ans, maxi);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}
void ipgraph(int n, int m) {
  int i, u, v;
  while (m--) {
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
}
void dfs(int u, int par) {
  for (int v : g[u]) {
    if (v == par) continue;
    dfs(v, u);
  }
}

```