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: Output Limit Exceeded)

#include <stdio.h>

int main() {
    long long int n, ans, sum, A[200005];
    int i,j;
    scanf("%lld", &n);
    for(i=0;i<n;i++) {
        scanf("%lld", A+i);
        if(A[i] >= 500000004) A[i] = A[i] - 1000000007;
    }
    
    ans = 0;
    sum = 0;

    for(i=0; i<n-1; i++) {
        for(j=i+1;j<n;j++) {
            sum = (A[i] * A[j]) % 1000000007;
            if (sum < 0) sum += 1000000007;
            ans += sum;
            ans = ans % 1000000007;
            printf("%lld\n",sum);
        }
    }

    
    printf("%lld",ans);
    return 0;
}