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)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|s| s.parse().ok().unwrap())
        .collect()
}

fn main() {
    let n = read::<usize>();
    let mut ls = read_vec::<usize>();

    ls.sort();
    ls.reverse();



    let mut count = 0;

    for i in 0..n {
        let a = ls[i];
        for j in i+1..n {
            let b = ls[j];
            let mut lo = j+1;
            let mut hi = n-1;
            let threshold = a - b;
            let mut cur = n;

            while lo <= hi {
                let mid = (hi+lo) / 2;
                let c = ls[mid];
                match c {
                    c if threshold < c => {
                        cur = mid;
                        lo = mid+1;
                    },
                    _ => {
                        hi = mid-1;
                    }
                }

            }

            if cur != n {
                count += cur - j;
            }

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