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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::Read;

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();

    let answer = solve(&buf);

    println!("{}", answer);
}

fn solve(input: &str) -> String {
    let mut iterator = input.split_whitespace();

    let n: usize = iterator.next().unwrap().parse().unwrap();
    let mut l: Vec<usize> = (0..n)
        .map(|_| iterator.next().unwrap().parse().unwrap())
        .collect();

    l.sort();

    println!("{:?}", l);

    let mut count = 0;
    for i in 0..(n - 2) {
        for j in (i + 1)..(n - 1) {
            let a = l[i];
            let b = l[j];

            if !can_triangle(a, b, l[j + 1]) {
                continue;
            }

            if can_triangle(a, b, l[n - 1]) {
                count += n - 1 - j;
                continue;
            }

            let mut left = j + 1;
            let mut right = n - 1;
            while left + 1 < right {
                let mid = (right + left) / 2;
                if can_triangle(a, b, l[mid]) {
                    left = mid;
                } else {
                    right = mid;
                }
            }
            count += left - j;
        }
    }
    return count.to_string();
}

fn can_triangle(a: usize, b: usize, c: usize) -> bool {
    return a < b + c && b < c + a && c < a + b;
}