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)

/*
 * main.c
 *
 *  Created on: 2020/03/28
 *      Author: family
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define MAX(a,b) (a > b ? a : b)
#define MIN(a,b) (a > b ? b : a)
typedef long long int ll;
typedef unsigned long long int ull;

int sort_inc(const void *a, const void *b) { return (*(int*)a - *(int*)b);}
int sort_dec(const void* a, const void* b) { return (*(int*)b - *(int*)a);}
int sort_dec_ll(const void *a, const void *b) {
    ll da = *(ll*)a, db = *(ll*)b;    int val = 0;
    if(da > db) { val = -1; }
    else if (da == db) { val = 0; }
    else { val = 1; }
    return val;
}
int sort_inc_ll(const void *a, const void *b) {
    ll da = *(ll*)a, db = *(ll*)b;    int val = 0;
    if(da > db) { val = 1; }
    else if (da == db) { val = 0; }
    else { val = -1; }
    return val;
}
int sort_dic(const void *a, const void *b) {
    char *pa = (char *)a;    char *pb = (char *)b;    int i = 0, val = 0, N = 10;
    for (i = 0; i < N; i++) {
    	char da = pa[i], db = pb[i];
    	if (da == db) continue;
  		if (da > db) val = 1; else val = -1;
        break;
    }
    return val;
}

void yesno(int ans) {
	if (ans == 1) {	printf("Yes\n");
	} else {		printf("No\n");	}
	return;
}

void okng(int ans) {
	if (ans == 1) { printf("OK\n");
	} else { 		printf("NG\n");	}
	return;
}

#define FIX_VAL (1000000000 + 7)
ll A[200002] = {0};
ll add[200002] = {0};
int main()
{
	int N = 0;
	scanf("%d", &N);
	int i = 0;
	for (i = 0; i < N; i++) {
		scanf("%lld", &A[i]);
	}

	int j = 0;
	ll prev = 0;
	for (j = N-1; j > 0; j--) {
		add[j] += A[j] + prev;
		prev = add[j];
		//printf("%lld A=%lld\n", add[j], A[j]);
	}


	ll ans = 0;
	for (i = 0; i < N-1; i++) {
		if (A[i] == 0) {
		    //
		} else {
			ans = ((A[i] * add[i+1])+ans) % FIX_VAL;
		}
	}
	printf("%lld\n", ans);
	return 0;
}