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: Accepted)

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

int compare_int(const void *a,const void *b){
		return *(int*)a-*(int*)b;
}

struct tm{
		int a;
		int n;
};

int main(){
		int n,i,c=0,l[2001],j;
		long long ans=0;
		scanf("%d",&n);
		for(i=0;i<n;i++)scanf("%d",&l[i]);
		qsort(l,n,sizeof(int),compare_int);
	/*	struct tm ll[1010];
		ll[0].a=l[0];
		ll[0].n=1;
		for(i=1;i<n;i++){
				if(l[i]!=l[i-1]){
						ll[++c].a=l[i];
						ll[c].n=0;
				}
				ll[c].n++;
		}c++;
		for(i=0;i<c;i++){
				for(j=i+1;j<c;j++){
						int r=c,le=j,q=0,k;
						while(r-le>1){
								int mid=(r+le)/2;
								if(ll[mid].a>=ll[i].a+ll[j].a)r=mid;
								else le=mid;
						}
						for(k=j+1;k<r;k++)q+=ll[k].n;
						ans+=(q*ll[i].n*ll[j].n);
						ans+=ll[i].n*ll[j].n*(ll[j].n-1)/2;
						if(ll[i].a*2>ll[j].a)ans+=ll[j].n*ll[i].n*(ll[i].n-1)/2;
				}
		}*/
		for(i=0;i<n;i++){
				for(j=i+1;j<n;j++){
						int r=n,le=j,q=0;
						while(r-le>1){
								int mid=(r+le)/2;
								if(l[mid]>=l[i]+l[j])r=mid;
								else le=mid;
						}
						ans+=le-j;
				}
		}
		printf("%lld\n",ans);
		return 0;
}