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.
//Bismillahir Rahmanir Rahim
//In the name of Allah the Most Merciful
//Astagfirullahil laji Rabbi Minkulli Jamio Watubi elai
#include<bits/stdc++.h>
using namespace std;
#define int long long int
#define double long double

#define ha cout<<"YES"<<endl;
#define na cout<<"NO"<<endl;
#define hm cout<<"Yes"<<endl;
#define np cout<<"No"<<endl;
#define G int tst;cin>>tst; while(tst--){
#define MX 10000008
#define suru ios::sync_with_stdio(false); cin.tie(0);
#define cs int t;cin>>t;while(t--)
#define bddao cout<<endl;
int i,j,k;
int r,s,c;
main()
{ int t;cin>>t;while(t--){
     int n;
     cin>>n;
     int a[n];
     int mx=0;
     int p=0;
     a[0]=0;
     for(i=1;i<=n;i++)
     {
         cin>>a[i];
          if(mx<a[i])
          {
              mx=a[i];
              p=i;
          }
     }
     int h1=mx;
     int h2=max(a[p-1],a[p+1]);
     cout<<h1*h2<<endl;

}
}
