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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
int compare_int(const void *a,const void *b){
  return *(int *)a-*(int *)b;
}
int main(void)
{
  int n,i,j;
  int left,right,mid,value,ans=0;
  scanf("%d",&n);
  int l[n];
  for(i=0;i<n;i++){
    scanf("%d",&l[i]);
  }
  qsort(l,n,sizeof(int),compare_int);
  for(i=0;i<n-2;i++){
    for(j=i+1;j<n-1;j++){
      left=j+1;
      right=n-1;
      value=l[i]+l[j];

      while(left<right){
        mid=(left+right)/2;
        if(l[mid]==value){
          break;
        } else if(l[mid]<value){
          left=mid+1;
        } else {
          right=mid-1;
        }
      }
      /*if(l[mid]<value){
        if(l[mid+1]<value){
          ans+=mid-j+1;
          printf("mid-j+1=%d\n",mid-j+1);
          //mid+1がl[i]+l[j]<l[k]をみたすkの最大値
        }else{
          ans+=mid-j;
          printf("mid-j=%d\n",mid-j);
          //midがl[i]+l[j]<l[k]をみたすkの最大値
        }
      }else{
        ans+=mid-j-1;
        printf("mid-j-1=%d\n",mid-j-1);
        //mid-1がl[i]+l[j]<l[k]をみたすkの最大値
      }
    }*/
      if(l[left]<value){
        if(l[left+1]<value){
          ans+=left-j+1;
        }else{
          ans+=left-j;
        }
      }else{
        ans+=left-j-1;
      }
    }
  }
  printf("%d",ans);
  return 0;
}