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 <stdlib.h>
#include <string.h>

int main(void)
{
  long long int n,a[200001],sum=0;

  scanf("%lld", &n);
  
  for(int i = 0;i < n;i++)
    scanf("%lld",&a[i]);
  
  for(int i = 0;i < n;i++){
    for(int j = i+1;j < n;j++){
      sum += a[i]*a[j];
    }
  }
  
  printf("%lld", sum%(1000000000+7));
  
  return 0;
}