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: Wrong Answer)

#include <stdio.h>
#define N 400000
#define mod 1000000007
int main(int argc, char *argv[])
{
	long long a[N]={0},n,i,sum=0;
	scanf("%lld",&n);
	for(i=0;i<n;i++){
		scanf("%lld",&a[i]);
		sum+=a[i];
	}
	long long s=0,dum;
	for(i=0;i<=n-2;i++){
		sum-=a[i];
		dum=sum;
		dum%=mod;
		s+=(dum*a[i])%mod;
	}
	printf("%lld\n",s);
}