#include <bits/stdc++.h>
const int MAX = (const int)(1e5 + 11);
using namespace std;
static void solve(std::int32_t test) {
  vector<int64_t> A(MAX);
  int T;
  cin >> T;
  for (int t = 0; t < T; ++t) {
    int N;
    cin >> N;
    for (int i = 0; i < N; ++i) cin >> A[i];
    int64_t res = 0;
    for (int i = 0; i < N; ++i) {
      for (int j = 1; j <= 10; ++j) {
        int64_t mx = 0;
        int64_t mn = 1e9;
        int cnt = 0;
        for (int k = i; k <= i + j; ++k) {
          if (0 <= k && k < N) {
            mx = max(mx, A[k]);
            mn = min(mn, A[k]);
            cnt++;
          }
        }
        if (cnt > 1) res = max(res, mx * mn);
      }
    }
    cout << res << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << setprecision(12) << fixed;
  solve(0);
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

You are given n integers a_1, a_2, …, a_n. Find the maximum value of max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r) over all pairs (l, r) of integers for which 1 ≤ l < r ≤ n.

Input

The first line contains a single integer t (1 ≤ t ≤ 10 000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5).

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

It is guaranteed that the sum of n over all test cases doesn't exceed 3 ⋅ 10^5.

Output

For each test case, print a single integer — the maximum possible value of the product from the statement.

Example

Input


4
3
2 4 3
4
3 2 3 1
2
69 69
6
719313 273225 402638 473783 804745 323328


Output


12
6
4761
381274500335

Note

Let f(l, r) = max(a_l, a_{l + 1}, …, a_r) ⋅ min(a_l, a_{l + 1}, …, a_r).

In the first test case, 

  * f(1, 2) = max(a_1, a_2) ⋅ min(a_1, a_2) = max(2, 4) ⋅ min(2, 4) = 4 ⋅ 2 = 8. 
  * f(1, 3) = max(a_1, a_2, a_3) ⋅ min(a_1, a_2, a_3) = max(2, 4, 3) ⋅ min(2, 4, 3) = 4 ⋅ 2 = 8. 
  * f(2, 3) = max(a_2, a_3) ⋅ min(a_2, a_3) = max(4, 3) ⋅ min(4, 3) = 4 ⋅ 3 = 12. 



So the maximum is f(2, 3) = 12.

In the second test case, the maximum is f(1, 2) = f(1, 3) = f(2, 3) = 6.

### Response

```cpp
#include <bits/stdc++.h>
const int MAX = (const int)(1e5 + 11);
using namespace std;
static void solve(std::int32_t test) {
  vector<int64_t> A(MAX);
  int T;
  cin >> T;
  for (int t = 0; t < T; ++t) {
    int N;
    cin >> N;
    for (int i = 0; i < N; ++i) cin >> A[i];
    int64_t res = 0;
    for (int i = 0; i < N; ++i) {
      for (int j = 1; j <= 10; ++j) {
        int64_t mx = 0;
        int64_t mn = 1e9;
        int cnt = 0;
        for (int k = i; k <= i + j; ++k) {
          if (0 <= k && k < N) {
            mx = max(mx, A[k]);
            mn = min(mn, A[k]);
            cnt++;
          }
        }
        if (cnt > 1) res = max(res, mx * mn);
      }
    }
    cout << res << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cout << setprecision(12) << fixed;
  solve(0);
  return 0;
}

```