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: Runtime Error)

#include <stdio.h>
#include <math.h>
#define DATANUMBER 30000
 
int main(void) {
	long long  N,i,j,sum;
	long long A[DATANUMBER+1];
 
	
	scanf("%lld",&N);
	for(i=1;i<=N;i++){
		scanf("%lld",&A[i]);
	}
	sum=0;
	for(i=1;i<=N;i++){
		for(j=i+1;j<=N;j++){
			A[i]=A[i]%(long long)(pow(10,9)+7);
			A[j]=A[j]%(long long)(pow(10,9)+7);
			sum+=(A[i]*A[j])%(long long)(pow(10,9)+7);
		}
	}
	sum=sum%(long long)(pow(10,9)+7);
	printf("%lld",sum);
	return 0;
 
}