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>
#include <vector>
#include <cmath>
#define REP(i, n) for (int i = 0; i < (int)(n); ++i)
#define REP2(i, m, n) for (int i = (m); i < (int)(n); ++i)
#define REPR(i, n) for (int i = (n)-1; i >= 0; --i)
#define REPR2(i, m, n) for (int i = (n)-1; i >= (m); --i)
#define REPx(x, a) for(auto x : a)
#define ALL(a) a.begin(), a.end()
#define SORT(a) sort(ALL(a))
#define SORTR(a) sort(ALL(a), greater<int>())
#define REVERSE(a) reverse(ALL(a))
#define bit_search(bit, n) REP(bit, 1<<(n))
#define bit_check(bit, i) (bit>>(i)) & 1
#define setpre(n) fixed << setprecision((n))
#define UNIQUE(a) do {SORT(a); (a).erase(unique(ALL(a)), (a).end());} while(0)
#define MAX(a) *max_element(ALL(a))
#define MIN(a) *min_element(ALL(a))
#define bisect_left(a, x) lower_bound(ALL(a), (x)) - a.begin()
#define bisect_right(a, x) upper_bound(ALL(a), (x)) - a.begin()
#define INPUT(a) REP(i, a.size()) cin >> a[i];
#define INPUTP(a) REP(i, a.size()) cin >> a[i].first >> a[i].second;
#define ENDL cout << endl;
using namespace std;
using LL = long long;
using LD = long double;
using PII = pair<int, int>;
using VPII = vector<PII>;
using PLL = pair<LL, LL>;
using VPLL = vector<PLL>;
using VI = vector<int>;
using VVI = vector<VI>;
using VLL = vector<LL>;
using VVLL = vector<VLL>;
using VC = vector<char>;
using VS = vector<string>;
using VB = vector<bool>;
const int INF = 1e9;
const LL MOD = 1e9 + 7;

template<class T> inline bool chmin(T &a, T b){if (a > b) {a = b; return true;} return false;}
template<class T> inline bool chmax(T &a, T b){if (a < b) {a = b; return true;} return false;}

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  VI d(n);
  INPUT(d);
  int ans = 0;
  REP(i, n-1) REP2(j, i+1, n) ans += d[i]*d[j];
  cout << ans << endl;

  return 0;
}