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)

#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(dead_code)]

fn main() {
    let N: usize = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.trim().parse().unwrap()
    };
    let L: Vec<i64> = {
        let mut line: String = String::new();
        std::io::stdin().read_line(&mut line).unwrap();
        line.split_whitespace()
            .map(|x| x.parse().unwrap())
            .collect()
    };

    let S = {
        let mut S = L.clone();
        S.sort();
        S
    };
    fn bis<P>(ok: i64, ng: i64, p: P) -> i64
        where P: Fn(i64) -> bool {
        let mid = (ok + ng) / 2;
        if (ok - ng).abs() == 1 {
            ok
        } else if p(mid) {
            bis(mid, ng, p)
        } else {
            bis(ok, mid, p)
        }
    }
    // println!("{:?}", S);
    let ans = (2..N).map(|i|
        (0..(i - 1)).map(|j| {
            let k = bis(i as i64, j as i64, |k| S[i] < S[j] + S[k as usize]) as usize;
            // println!("{} {} {}", i, j, k);
            if k > j { i - k } else { 0 }
        }).sum::<usize>()
    ).sum::<usize>();

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