#include <bits/stdc++.h>
using namespace std;
void seive(long long);
bool cmp(long long, long long);
vector<bool> prime(200002, true);
void solve() {
  long long n, i, j, k;
  cin >> n;
  long long arr[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
  }
  if (n == 2) {
    cout << arr[0] * arr[1] << endl;
    return;
  }
  long long maximum = INT_MIN, prod = 1, second = 0;
  for (i = 0; i < n; i++) {
    if (i == 0) {
      prod = max(prod, arr[i] * arr[i + 1]);
    } else if (i == n - 1) {
      prod = max(prod, arr[i] * arr[i - 1]);
    } else {
      prod = max(prod, max(arr[i - 1] * arr[i], arr[i + 1] * arr[i]));
    }
  }
  cout << prod << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
void seive(long long mx) {
  prime[0] = prime[1] = false;
  for (long long i = 2; i * i <= mx; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= mx; j += i) {
        prime[j] = false;
      }
    }
  }
}
bool cmp(long long x, long long y) {
  if (x > y)
    return true;
  else
    return false;
}

### Prompt

Generate 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>
using namespace std;
void seive(long long);
bool cmp(long long, long long);
vector<bool> prime(200002, true);
void solve() {
  long long n, i, j, k;
  cin >> n;
  long long arr[n];
  for (i = 0; i < n; i++) {
    cin >> arr[i];
  }
  if (n == 2) {
    cout << arr[0] * arr[1] << endl;
    return;
  }
  long long maximum = INT_MIN, prod = 1, second = 0;
  for (i = 0; i < n; i++) {
    if (i == 0) {
      prod = max(prod, arr[i] * arr[i + 1]);
    } else if (i == n - 1) {
      prod = max(prod, arr[i] * arr[i - 1]);
    } else {
      prod = max(prod, max(arr[i - 1] * arr[i], arr[i + 1] * arr[i]));
    }
  }
  cout << prod << endl;
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
void seive(long long mx) {
  prime[0] = prime[1] = false;
  for (long long i = 2; i * i <= mx; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= mx; j += i) {
        prime[j] = false;
      }
    }
  }
}
bool cmp(long long x, long long y) {
  if (x > y)
    return true;
  else
    return false;
}

```