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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        ArrayList<Integer> ll = new ArrayList<>();
        for(int i = 0; i < n; i++){
            ll.add(sc.nextInt());
        }
        Collections.sort(ll);
        long ans = 0;
        for(int i = n-1; i >= 0; i--){
            int big = ll.get(i);
            for(int j = i-1; j >= 0; j--){
                int small = ll.get(j);
                int res = beamSearch(ll,big-small+1);
                //System.out.println(i + " " + j + " " + (j > res ? j-res : 0));
                ans += (j > res) ? j-res : 0;
            }
        }
        System.out.println(ans);
    }
    
    public static int beamSearch(ArrayList<Integer> ll, int k){
        int left = 0;
        int right = ll.size();
        while(right - left > 1){
            int mid = (right + left) / 2;
            if(ll.get(mid) > k){
                right = mid;
            }else{
                left = mid;
            }
        }
        return left;
    }
}