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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for (int (i)=0;(i)<(n);i++)
#define INF 1001001001
#define LLINF 1001001001001001001
#define MOD 1000000007
#define FOUT(n, dist) cout<<fixed<<setprecision(n)<<(dist);
#define BR cout<<endl;
#define LEN(s) s.length();
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define ALL(v) v.begin(), v.end()
typedef long long ll;
const ll mod=1000000007;

ll powmod(ll x,ll y){
    ll res=1;
    for(ll i=0;i<y;i++){
        res=res*x%mod;
    }
    return res;
}

double expect(int x){
    double kitai,tem=0;
    for(int i=1;i<=x;i++){
        tem += i;
    }
    kitai = tem/x;
    cout << kitai << endl;
    return kitai;
    
}

int main(){
    int n;
    cin >> n;

    ll a[n];
    rep(i,n)cin >> a[i];

    ll sum = 0, ans = 0;

    for(int i = 0; i < n; i++){
        for(int j = i + 1; j < n; j++){
            sum += a[i] * a[j];
            sum = sum % mod;
        }
    }

    cout << sum << endl;

    return 0;
}