#include <iostream>
#include <bits/stdc++.h>
using namespace std;
 
int main() {
     ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	int num;
	cin >> num;										// Reading input from STDIN
	while(num--)
	{
	  
	    int n;
	    cin>>n;
	    int A[n];
	    int ma=0;
	    long long sum=0;
	    for(int i=0;i<n;++i)
	    {
	    cin>>A[i];
	    if(ma<A[i])
	    ma=A[i];
	    sum=sum+A[i];
	    }
	    sort(A,A+n);
	    long long res=0;
	    for(int i=0;i<n-1;++i)
	    {
	        for(int j=i+1;j<n;++j)
	        {
	            res=res+abs(A[i]-A[j]);
	        }
	      
	    }
	    res=res*ma;
	    cout<<res%1000000007<<"\n";
	}
}
 

Problem Description

Andrew is very fond of Maths.He has N boxes with him,in each box there is some value which represents the Strength of the Box.The ith box has strength A[i]. He wants to calculate the Overall Power of the all N Boxes.

Overall Power here means Sum of Absolute Difference of the strengths of the boxes(between each pair of boxes) multiplied by the Maximum strength among N boxes. Since the Overall Power could be a very large number,output the number modulus 10^9+7(1000000007).

Input

First line of the input contains the number of test cases T. It is followed by T test cases. Each test case has 2 lines. First line contains the number of boxes N. It is followed by a line containing N elements where ith element is the strength of Andrew’s ith box.

Output

For each test case, output a single number, which is the Overall Power for that testcase.

Constraints

1<=T<= 10

2<=N<=10^5

0<=A[i]<=10^9

• Test Case 1

  Input (stdin)

  1
  3
  3 1 2
  

  Expected Output

  12

• Test Case 2

  Input (stdin)

  2
  2
  1 2
  5
  4 5 3 1 2
  

  Expected Output

  2
  100