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 (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <limits.h>
#include <time.h>
int main(){
    long long i,j,n,ans=0,a[200020]={},sum[200020]={};
    scanf("%lld",&n);
    for(i=0; i<n; i++){
        scanf("%lld",&a[i]);
        sum[0]+=a[i];
    }
    for(i=1; i<n; i++){
        sum[i]=sum[i-1]-a[i-1];
    }
    for(i=0; i<n; i++){
        sum[i]=sum[i]%1000000007;
    }
    for(i=0; i<n-1; i++){
        ans+=a[i]*sum[i+1];
        ans=ans%1000000007;
    }
    printf("%lld",ans);
    return 0;
}