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 <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp>
using namespace std; 

// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

# define ll long long int
#define M 1000000007
#define pb push_back
#define ss second
#define ff first
#define inf 1000000000000000


 

													//nestedcode



int main(){
	ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n,i,ans = 0;
    cin>>n;
    ll a[n+5], dp[n+5];
    for(i=0;i<n;i++){
    	cin>>a[i];
    	if(i == 0)
    	dp[i] = a[i];
    	else
    	dp[i] = dp[i-1] + a[i];
	}
	for(i=1;i<n;i++){
		ans += (dp[i-1]*a[i])%M;
		ans %= M;
	}
	cout<<ans<<endl;
}