#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return a;
  else {
    if (n % 2 == 0)
      return (power(a, n / 2) % 1000000007 * power(a, n / 2) % 1000000007) %
             1000000007;
    else
      return (power(a, n / 2) % 1000000007 * power(a, n / 2) % 1000000007 * a) %
             1000000007;
  }
}
bool poweroftwo(int x) { return x && (!(x & (x - 1))); }
long long int cmpi(pair<long long int, long long int> a,
                   pair<long long int, long long int> b) {
  return a.first < b.first;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int search(long long int a[], long long int n, long long int x) {
  long long int l = 0, r = n - 1;
  if (x < a[l] || x > a[r]) return -1;
  while (l <= r) {
    long long int mid = l + (r - l) / 2;
    if (a[mid] == x) return 1;
    if (a[mid] < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return -1;
}
long long int isprime(long long int x) {
  long long int flag = 0;
  for (long long int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      flag++;
      break;
    }
  }
  if (flag) return 1;
  return 0;
}
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int ans = LLONG_MIN;
  for (long long int i = 1; i < n - 1; i++) {
    ans = max(ans, max(min(a[i], a[i - 1]) * max(a[i], a[i - 1]),
                       min(a[i], a[i + 1]) * max(a[i], a[i + 1])));
  }
  ans = max(ans, a[0] * a[1]);
  ans = max(ans, a[n - 1] * a[n - 2]);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) solve();
}

### 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;
long long int power(long long int a, long long int n) {
  if (n == 0)
    return 1;
  else if (n == 1)
    return a;
  else {
    if (n % 2 == 0)
      return (power(a, n / 2) % 1000000007 * power(a, n / 2) % 1000000007) %
             1000000007;
    else
      return (power(a, n / 2) % 1000000007 * power(a, n / 2) % 1000000007 * a) %
             1000000007;
  }
}
bool poweroftwo(int x) { return x && (!(x & (x - 1))); }
long long int cmpi(pair<long long int, long long int> a,
                   pair<long long int, long long int> b) {
  return a.first < b.first;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int search(long long int a[], long long int n, long long int x) {
  long long int l = 0, r = n - 1;
  if (x < a[l] || x > a[r]) return -1;
  while (l <= r) {
    long long int mid = l + (r - l) / 2;
    if (a[mid] == x) return 1;
    if (a[mid] < x)
      l = mid + 1;
    else
      r = mid - 1;
  }
  return -1;
}
long long int isprime(long long int x) {
  long long int flag = 0;
  for (long long int i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      flag++;
      break;
    }
  }
  if (flag) return 1;
  return 0;
}
void solve() {
  long long int n;
  cin >> n;
  long long int a[n];
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long int ans = LLONG_MIN;
  for (long long int i = 1; i < n - 1; i++) {
    ans = max(ans, max(min(a[i], a[i - 1]) * max(a[i], a[i - 1]),
                       min(a[i], a[i + 1]) * max(a[i], a[i + 1])));
  }
  ans = max(ans, a[0] * a[1]);
  ans = max(ans, a[n - 1] * a[n - 2]);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) solve();
}

```