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: Accepted)

#include<bits/stdc++.h>
using namespace std; 
 
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize ("-ffloat-store")  
#pragma GCC optimize ("-fno-defer-pop")
 
typedef long long int ll; 
typedef long double ld; 

ll mod = 1e9+7;

ll add(ll a, ll b){
	a = a%mod;
	b = b%mod;
	return (mod+(a+b)%mod)%mod;
}

ll mul(ll a, ll b){
	a = a%mod;
	b = b%mod;
	return (mod+(a*b)%mod)%mod;
}

int main(){
	
	std::ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	
	ll n;
	cin>>n;
	
	ll arr[n];
	ll su = 0;
	for(ll i=0;i<n;i++){
		cin>>arr[i];
		su += arr[i];
	}
	ll ans = 0;
	for(ll i=0;i<n;i++){
		su-=arr[i];
		ans = add(ans, mul(arr[i], su));
	}	
	cout<<ans<<endl;
}