Problem:
Score : 400 points

Problem Statement

Takahashi has N sticks that are distinguishable from each other. The length of the i-th stick is L_i.

He is going to form a triangle using three of these sticks. Let a, b, and c be the lengths of the three sticks used. Here, all of the following conditions must be satisfied:

a < b + c

b < c + a

c < a + b

How many different triangles can be formed? Two triangles are considered different when there is a stick used in only one of them.

Constraints

All values in input are integers.

3 \leq N \leq 2 \times 10^3

1 \leq L_i \leq 10^3

Input

Input is given from Standard Input in the following format:

N
L_1 L_2 ... L_N

Constraints

Print the number of different triangles that can be formed.

Sample Input 1

4
3 4 2 1

Sample Output 1

1

Only one triangle can be formed: the triangle formed by the first, second, and third sticks.

Sample Input 2

3
1 1000 1

Sample Output 2

0

No triangles can be formed.

Sample Input 3

7
218 786 704 233 645 728 389

Sample Output 3

23

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

#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)

int main(){
  int N = 0;
  cin >> N;
  vector<int> L(N);
  rep(i, N){
    cin >> L.at(i);
  }
  vector<int> b(1001, 0);
  rep(i, N){
    b.at(L.at(i))++;
  }
  int64_t cnt = 0;
  int64_t temp1, temp2, temp3 = 0;
  rep(i, 1001){
    if(b.at(i) > 0){
      temp1 = b.at(i);
      b.at(i)--;
      rep2(j, i, 1001){
        if(b.at(j) > 0){
          temp2 = b.at(j);
          b.at(j)--;
          rep2(k, j, 1001){
            if(b.at(k) > 0){
              temp3 = b.at(k);
              if(i < j+k && j < k+i && k < i+j){
                if(i == j && j == k){
                  cnt += temp1 * temp2 * temp3 / 6;
                }
                else if(i == j || j == k || k == i){
                  cnt += temp1 * temp2 * temp3 / 2;
                }
                else{
                cnt += temp1 * temp2 * temp3;
                }
              }
            }
          }
          b.at(j)++;
        }
        
      }
      b.at(i)++;
    }
  }
  
  cout << cnt << endl;
}