#include <bits/stdc++.h>
using namespace std;
long long int MOD = 1E9 + 7;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int gcd_ex(long long int a, long long int b, long long int& x,
                     long long int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int x1, y1;
  long long int g = gcd_ex(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return g;
}
long long int multi_mod_inv(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcd_ex(a, m, x, y);
  if (g != 1) {
    return -1;
  }
  x = (x % m + m) % m;
  return x;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  if (multi_mod_inv(b, m) != -1) {
    return (a % m * multi_mod_inv(b, m) % m) % m;
  } else {
    return -1;
  }
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int power(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int p = power(a, n / 2);
  p = (p * p) % MOD;
  if (n % 2)
    return (p * a) % MOD;
  else
    return p % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int test_case = 1;
  cin >> test_case;
  for (long long int test = 1; test <= test_case; test++) {
    long long int n;
    cin >> n;
    long long int ans = 0;
    vector<long long int> a(n);
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (long long int i = 1; i < n; i++) {
      ans = max(ans, a[i] * a[i - 1]);
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long int MOD = 1E9 + 7;
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int gcd_ex(long long int a, long long int b, long long int& x,
                     long long int& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long int x1, y1;
  long long int g = gcd_ex(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return g;
}
long long int multi_mod_inv(long long int a, long long int m) {
  long long int x, y;
  long long int g = gcd_ex(a, m, x, y);
  if (g != 1) {
    return -1;
  }
  x = (x % m + m) % m;
  return x;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  if (multi_mod_inv(b, m) != -1) {
    return (a % m * multi_mod_inv(b, m) % m) % m;
  } else {
    return -1;
  }
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int power(long long int a, long long int n) {
  if (n == 0) return 1;
  long long int p = power(a, n / 2);
  p = (p * p) % MOD;
  if (n % 2)
    return (p * a) % MOD;
  else
    return p % MOD;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int test_case = 1;
  cin >> test_case;
  for (long long int test = 1; test <= test_case; test++) {
    long long int n;
    cin >> n;
    long long int ans = 0;
    vector<long long int> a(n);
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    for (long long int i = 1; i < n; i++) {
      ans = max(ans, a[i] * a[i - 1]);
    }
    cout << ans << '\n';
  }
  return 0;
}

```