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>
#include<math.h>

int main(void){
    int N;
    scanf("%d",&N);
    long long A[N];
    long long sum = 0, answer;
    for(int i = 0; i < N; i++){
        scanf("%d",&A[i]);
    }
    for(int i = 0; i < N-1; i++){
        for(int j = i+1; j < N; j++){
            if(A[i]*A[j] > 2*pow(10,9)){
                answer = A[i]*A[j] % (long long)(pow(10,9)+7);
            }
            sum += answer;
            
            if(sum > (pow(10,9)+7)){
                sum %= (long long)(pow(10,9)+7);
            }
           // printf("%ld, %ld\n",answer,sum);
        }
    }
    printf("%ld\n",sum%(long)(pow(10,9)+7));
    return 0;
}