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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>


int compare_int(const void *a, const void *b)
{
    return *(int*)a - *(int*)b;
//    return *(long long*)a - *(long long*)b;
}
//  qsort(A, b, sizeof(int), compare_int);

long long  factorial(long long n) {
    if (n > 0) {
        return n * factorial(n - 1);
    } else {
        return 1;
    }
}

int min(int a, int b) {
    if (a > b)
        return b;
    else
        return a;
}

int distance(int a, int b) {
  return abs(a-b)*abs(a-b);
}

//int ()
/*
int triangle(int a, int b, int c){
  if(a<b+c){
    if(b<c+a){
      if(c<a+b){
        printf("1\n");
        return 1;
      }
    }
  }
  printf("1\n");
  return 0;
}
*/

int main()
{
  int  a ,b, c, d, e, X, Y ,r;
//  long long  a, b, c, d, e, X, Y;
  //  double  c;
//  int  A[200003]={};
  int  A[100003]={};
  int  B[100003]={};
  int  C[100003][11]={};

  int  flg=0;

//  long long  a, b, c, d, X, Y;
//  char s[101][101]={};
  char s[100001]={};
  char t[101]={};
//  int N, M;
//  int A[1000000001]={};
//  long long  C[101]={};
  int i,j,k;
  long long p,q;
//  int flag=0;
//  int tmp=0;
//  char tmp[101]={};
//  long long min=10000000000;
//  long long min=100000001;
  int len=0;
  int l=0;
//  int r=1000001;
  int cnt=0;
  int num=0;
  int num1=0;
  int num2=0;
  int ans=0;
  int max=0;
  int sum=0;
  int addr=0;
  int ptn=0;
  long long  tmp=0;
//  bool on_off[100]={};
  
//  char alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

  //  a=0;
  scanf("%d", &a);
  
  for(i=0;i<a;i++){
    scanf("%d", &A[i]);
  }
 
  for(i=0;i<a-2;i++){
    for(j=i+1;j<a-1;j++){
      for(k=j+1;k<a;k++){

        if(A[i]<A[j]+A[k]){
          if(A[j]<A[k]+A[i]){
            if(A[k]<A[i]+A[j]){
              cnt++;
//              printf("1:i=%d i=%d i=%d\n", A[i], A[j], A[k]);
            }
          }
        }
      }
    }
  }

 
  printf("%d\n", cnt);
  
return 0;
}