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: Accepted)

import java.util.*;

public class Main implements Runnable {

    private static int MOD = 1_000_000_007;

    public static void main(String[] args) {
        // Run with 32MB stack
        Thread thread = new Thread(null, new Main(), "", 32 * 1024 * 1024);
        thread.start();
    }

    @Override
    public void run() {
        final Scanner scanner = new Scanner(System.in);
        solve(scanner);
    }

    static void solve(Scanner scanner) {
        int N = Integer.parseInt(scanner.next());
        int[] A = new int[N];
        for (int i = 0; i < N; i++) {
            A[i] = Integer.parseInt(scanner.next());
        }
        Arrays.sort(A);
        long ans = 0;
        for (int i = 0; i < N; i++) {
            for (int j = i + 1; j < N; j++) {
                int sum = A[i] + A[j];
                int index = binarySearch(A, j + 1, N, sum);
                if (index == N || A[index] == sum) {
                    index--;
                }
                ans += index - j;
            }
        }

        System.out.println(ans);
    }


    static int binarySearch(int[] A, int l, int r, int val) {
        if (A[r - 1] < val) {
            return r;
        }
        int left = l - 1, right = r;
        while (right - left > 1) {
            int mid = (left + right) >>> 1;
            if (A[mid] < val) {
                left = mid;
            } else {
                right = mid;
            }
        }
        return left;
    }
}