Problem:
Score : 300 points

Problem Statement

Given are N integers A_1,\ldots,A_N.

Find the sum of A_i \times A_j over all pairs (i,j) such that 1\leq i < j \leq N, modulo (10^9+7).

Constraints

2 \leq N \leq 2\times 10^5

0 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 \ldots A_N

Output

Print \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} A_i A_j, modulo (10^9+7).

Sample Input 1

3
1 2 3

Sample Output 1

11

We have 1 \times 2 + 1 \times 3 + 2 \times 3 = 11.

Sample Input 2

4
141421356 17320508 22360679 244949

Sample Output 2

437235829

Solution in C (Clang 10.0.0) (judge verdict: Accepted)

#include<stdio.h>

#define mod 1000000007

int main(void){
	int N;
	long long sum=0;
	long long ans=0;
	scanf("%d",&N);

	int A[N];

	for(int i=0;i<N;i++) scanf("%d",&A[i]);
    
	for(int i=0;i<N;i++){
		sum+=A[i];
		sum%=mod;
	}

    for(int i=0;i<N-1;i++){
		sum-=A[i];
		if(sum<0) sum+=mod;

		ans+=A[i]*sum;
		ans%=mod;
	}

	printf("%lld\n",ans);


	return 0;
}