#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long N = 5e5 + 10;
template <typename... T>
void debug(T&... args) {
  cout << "\n==================\n";
  ((cout << args << ", "), ...);
  cout << "\n==================\n";
}
template <typename T>
void debug(vector<T>& a) {
  cout << "\n==============\n";
  for (auto& e : a) cout << e << ' ';
  cout << "\n==============\n";
}
template <typename... T>
void put(T&... args) {
  ((cout << args << " "), ...);
}
void solve() {
  long long m, n;
  cin >> m >> n;
  vector<vector<long long> > p(m, vector<long long>(n));
  for (auto& e : p)
    for (auto& f : e) cin >> f;
  vector<pair<long long, long long> > ans(n, {-1, -1});
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (p[i][j] > ans[j].first) {
        ans[j].first = p[i][j];
        ans[j].second = i;
      }
    }
  }
  map<long long, long long> mp;
  for (auto& e : ans) mp[e.second]++;
  long long sol = 1e18;
  for (long long i = 0; i < n; i++) sol = min(sol, ans[i].first);
  if (mp.size() < n) {
    cout << sol << '\n';
    return;
  }
  sol = -1;
  for (long long i = 0; i < m; i++) {
    pair<long long, long long> p1;
    pair<long long, long long> p2;
    vector<pair<long long, long long> > gifts(n);
    for (long long j = 0; j < n; j++) {
      gifts[j].first = p[i][j];
      gifts[j].second = j;
    }
    sort(gifts.begin(), gifts.end());
    p1 = gifts[n - 1];
    p2 = gifts[n - 2];
    long long temp = 1e18;
    for (long long j = 0; j < n; j++) {
      if (p1.second == j)
        temp = min(temp, p1.first);
      else if (p2.second == j)
        temp = min(temp, p2.first);
      else
        temp = min(temp, ans[j].first);
    }
    sol = max(sol, temp);
  }
  cout << sol << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 long long mod = 998244353;
const long long N = 5e5 + 10;
template <typename... T>
void debug(T&... args) {
  cout << "\n==================\n";
  ((cout << args << ", "), ...);
  cout << "\n==================\n";
}
template <typename T>
void debug(vector<T>& a) {
  cout << "\n==============\n";
  for (auto& e : a) cout << e << ' ';
  cout << "\n==============\n";
}
template <typename... T>
void put(T&... args) {
  ((cout << args << " "), ...);
}
void solve() {
  long long m, n;
  cin >> m >> n;
  vector<vector<long long> > p(m, vector<long long>(n));
  for (auto& e : p)
    for (auto& f : e) cin >> f;
  vector<pair<long long, long long> > ans(n, {-1, -1});
  for (long long i = 0; i < m; i++) {
    for (long long j = 0; j < n; j++) {
      if (p[i][j] > ans[j].first) {
        ans[j].first = p[i][j];
        ans[j].second = i;
      }
    }
  }
  map<long long, long long> mp;
  for (auto& e : ans) mp[e.second]++;
  long long sol = 1e18;
  for (long long i = 0; i < n; i++) sol = min(sol, ans[i].first);
  if (mp.size() < n) {
    cout << sol << '\n';
    return;
  }
  sol = -1;
  for (long long i = 0; i < m; i++) {
    pair<long long, long long> p1;
    pair<long long, long long> p2;
    vector<pair<long long, long long> > gifts(n);
    for (long long j = 0; j < n; j++) {
      gifts[j].first = p[i][j];
      gifts[j].second = j;
    }
    sort(gifts.begin(), gifts.end());
    p1 = gifts[n - 1];
    p2 = gifts[n - 2];
    long long temp = 1e18;
    for (long long j = 0; j < n; j++) {
      if (p1.second == j)
        temp = min(temp, p1.first);
      else if (p2.second == j)
        temp = min(temp, p2.first);
      else
        temp = min(temp, ans[j].first);
    }
    sol = max(sol, temp);
  }
  cout << sol << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) solve();
}

```