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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		Integer[] bar = new Integer[n];
		for ( int i = 0; i < n; i++ ) {
			bar[i] = in.nextInt();
		}
		in.close();
		
		Arrays.sort(bar);
		int barmax = bar[n - 1];

		// i以上の長さのbarの個数＝lenNum[i]
		int[] lenNum = new int[barmax + 1];
		for ( int i = 0; i < n; i++ ) {
			lenNum[bar[i]]++;
		}
		for ( int i = barmax; 0 < i; i-- ) {
			lenNum[i - 1] += lenNum[i];
		}

		int cnt = 0;
		for ( int S = 0; S < n - 2; S++ ) {
			for ( int M = S + 1; M < n - 1; M++ ) {
				if ( bar[M] == barmax ) {
					cnt += (n - M) * (n - 1 - M) / 2;
					M = n;
				} else if ( barmax < bar[S] + bar[M] ) {
					cnt += lenNum[bar[M] + 1];
				} else {
					cnt += lenNum[bar[M] + 1] - lenNum[bar[S] + bar[M]];
				}
			}
		}
		System.out.println(cnt);
	}
}