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>
using namespace std;
#define MOD 1000000007
#define ll long long
#define ull unsigned long long
#define sii set <ll>
#define vi vector <ll>
#define popcount(x) __builtin_popcountll(x)
#define mii map <ll,ll>
#define vpi vector <pair <ll,ll> >
#define sz(c) (int)c.size()
#define fr first
#define ll long long
#define fastio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define sc second
#define pb push_back
#define mp make_pair
#define all(a) (a).begin(),(a).end()
#define mem0(a) memset(a,0,sizeof(a))
#define rep(i,a,n) for(ll i=a ; i<n ; i++)
#define ld long double
#define rall(a) (a).rbegin(),(a).rend()
#define INF 1000000000
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
 
 
 
ll power(ll b,ll e,ll m)
{
  if(b==0) return 0;
  if(e==0) return 1;
  if(e&1) return b*power(b*b%m,e/2,m)%m;
  return power(b*b%m,e/2,m);
}
ll power( ll b, ll e)
{
  if(b==0) return 0;
  if(e==0) return 1;
  if(e&1) return b*power(b*b,e/2);
  return power(b*b,e/2);
}  
bool isPowerOfTwo(ll x)
{
    // x will check if x == 0 and !(x & (x - 1)) will check if x is a power of 2 or not
    return (x && !(x & (x - 1)));
}


 void solution(ll a[],ll n)
 {
 	ll sum=0;
 	rep(i,0,n)
 	sum=(sum%MOD+a[i]%MOD)%MOD;

 	ll square = ((sum%MOD)*(sum%MOD))%MOD;

 	ll sum2=0;

 	rep(i,0,n)
 	{
 		sum2=((sum2%MOD)+((a[i]%MOD)*(a[i]%MOD))%MOD)%MOD;
 	}

 	// cout<<square<<"\n";
 	// cout<<sum2<<"\n";

 	ll ans= ((square%MOD - sum2%MOD)+MOD)%MOD;

 	cout<<ans/2;
 }


void solve()
{
	ll n;
	cin>>n;
	ll a[n];
	rep(i,0,n) cin>>a[i];
	solution(a,n);
}


int main()
{
	// freopen("input.txt","r",stdin);
	// freopen("output.txt","w",stdout);
  fastio
  int tt=1;
  // cin>>tt;
  while(tt--)
  {
    solve();
    cout<<"\n";
  }
  return 0;
}