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: Wrong Answer)

#include <algorithm>
#include<iostream>
#include<vector>
#include<list>
#include<map>
#include<set>
#include<string>

#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<limits.h>
using lll = long long;
using ull = unsigned long long;
using namespace std;


int main(){
  lll ii,jj,kk;
  vector<int> ret;
  lll n;
  vector<lll> l;
  cin >> n;
  l.resize(n);

  for(ii=0;ii<n;ii++){
    cin >> l[ii];
  }
  lll sum = 0;
  sort(l.begin(),l.end(),std::greater<lll>());
  //cout << "end sort" << endl;
  vector<lll> tmp;

  for(ii=0;ii<n-2;ii++){
    for(jj=ii+1;jj<n-1;jj++){

//      tmp.resize(n-(jj));
//
//      for(kk=0;kk<tmp.size()-1;kk++){
//        tmp[kk] = l[jj+1+kk];
//      }
      auto up1 = upper_bound(l.begin()+jj+1,l.end(),l[ii]-l[jj]-1);
      lll num = up1-l.begin();
      //cout << num << endl;
      sum += n - num;
//      for(kk=0;kk<tmp.size()-1;kk++){
//        if(l[jj]+tmp[kk]>l[ii]){
//          sum++;
//        }
//      }
      
    }
  }
  
  cout << sum << endl;
  return 0;
}