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: Accepted)

use std::io::{self, Read};

fn read_stdin() -> String {
    let mut buffer = String::new();
    io::stdin().read_to_string(&mut buffer).unwrap();
    buffer
}

fn parse_params(s: &str) -> Vec<i32> {
    let mut it = s.split_whitespace();
    let n = it.next().unwrap().parse().unwrap();

    let mut ret = Vec::with_capacity(n);

    for _ in 0..n {
        let l = it.next().unwrap().parse().unwrap();
        ret.push(l);
    }

    ret.sort();

    ret
}

fn count(v: &[i32], boundary: i32) -> usize {
    if v.is_empty() {
        return 0;
    }

    if boundary <= *v.first().unwrap() {
        return 0;
    }

    if *v.last().unwrap() < boundary {
        return v.len();
    }

    let half = v.len() / 2;
    let fst = &v[..half];
    let snd = &v[half..];

    count(fst, boundary) + count(snd, boundary)
}

fn main() {
    let buffer = read_stdin();
    let ls = parse_params(&buffer);

    let mut sum = 0;

    for i in 0..(ls.len() - 2) {
        for j in (i + 1)..(ls.len() - 1) {
            let boundary = ls[i] + ls[j];
            let ls = &ls[(j + 1)..];
            sum += count(ls, boundary);
        }
    }

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