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: Time Limit Exceeded)

/*
*          Name : chaitanya
*
*     Institute : ANITS,Visakhapatnam
*
*/
#include<cstdio>
#include<iostream>
#include<sstream>
#include<fstream>
#include<iomanip>
#include<algorithm>
#include<cmath>
#include<complex>
#include<cstring>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<bitset>
#include<numeric>
#include<limits>
#include<climits>
#include<cfloat>
#include<functional>
#include<iterator>
#include <random>
#define inf  1000000000
#define all(x) begin(x),end(x)
#define ll long long int
#define ff first
#define ss second
#define pb push_back
const ll mod = 1000000007;
#pragma GCC optimization ("O3")
const int N = 100005;
using namespace std;

int main()
{
 
  // freopen("input.txt","r",stdin);
  // freopen("output.txt","w",stdout);
  //ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
  int n,c = 0;
  cin >> n;
  vector <int> v(n);
  for (int i = 0; i < n; ++i)
  {
    cin >> v[i];
  }
  for (int i = 0; i < n; ++i)
  {
    for (int j = 0; j < n; ++j)
    {
      for (int k = 0; k < n; ++k)
      {
        if(i != j && j != k && k != i)
        {
          if(v[i]+v[j] > v[k] && v[i]+v[k] > v[j] && v[k]+v[j] > v[i])
            c++;
        }

      }
    }
  }
  cout << c/6;
  return 0;
}