Problem:
Score : 200 points

Problem Statement

It's now the season of TAKOYAKI FESTIVAL!

This year, N takoyaki (a ball-shaped food with a piece of octopus inside) will be served. The deliciousness of the i-th takoyaki is d_i.

As is commonly known, when you eat two takoyaki of deliciousness x and y together, you restore x \times y health points.

There are \frac{N \times (N - 1)}{2} ways to choose two from the N takoyaki served in the festival. For each of these choices, find the health points restored from eating the two takoyaki, then compute the sum of these \frac{N \times (N - 1)}{2} values.

Constraints

All values in input are integers.

2 \leq N \leq 50

0 \leq d_i \leq 100

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the sum of the health points restored from eating two takoyaki over all possible choices of two takoyaki from the N takoyaki served.

Sample Input 1

3
3 1 2

Sample Output 1

11

There are three possible choices:

Eat the first and second takoyaki. You will restore 3 health points.

Eat the second and third takoyaki. You will restore 2 health points.

Eat the first and third takoyaki. You will restore 6 health points.

The sum of these values is 11.

Sample Input 2

7
5 0 7 8 3 3 2

Sample Output 2

312

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<bits/stdc++.h>
#define F first
#define S second
#define endl '\n'
#define lcm( a, b ) ((a)/__gcd(a,b)*(b))
#define MP make_pair
#define pb push_back
#define all(x) (x).begin(),(x).end()

using namespace std;

typedef long long LL;
typedef pair<int, int> P;
typedef pair<int, P> iP;
typedef pair< LL, LL > LP;

const int INF = INT_MAX;
const int MIN = INT_MIN;
const LL LINF = LLONG_MAX;
const int MOD = 1000000000 + 7;
const int SIZE = 300005;

vector< int > Div( int n ) {
    vector< int > ret;
    for( int i = 1; i * i <= n; ++i ) {
        if( n % i == 0 ) {
            ret.pb( i );
            if( i * i != n ) ret.pb( n / i );

        }
    }
    sort( all( ret ) );
    return ret;
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    int res = 0;
    int d[n];
    for( int i = 0; i < n; ++i ) cin >> d[i];

    for( int i = 0; i < n; ++i ) {
        for( int j = i + 1; j < n; ++j ) {
            res += d[i] * d[j];

        }
    }

    cout << res << endl;
    return 0;
}