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(unused_macros)]
macro_rules! input {
    ( $($t:ty),* ) => {{
        let mut s = String::new();
        std::io::stdin().read_line(&mut s);
        let mut splits = s.trim().split_whitespace();
        ($( { splits.next().unwrap().parse::<$t>().unwrap() },)*)
    }}
}

#[allow(unused_macros)]
macro_rules! invec {
    ( $ t : ty ) => {{
        let mut s = String::new();
        match std::io::stdin().read_line(&mut s) {
            Ok(0) => Vec::<$t>::new(),
            Ok(n) => s
                .trim()
                .split_whitespace()
                .map(|s| s.parse::<$t>().unwrap())
                .collect::<Vec<$t>>(),
            Err(_) => Vec::<$t>::new(),
        }
    }};
}



#[allow(unused_must_use)]
#[allow(unused_variables)]
fn solve() {
    let (n, ) = input!(usize);
    let mut ls = invec!(u32);

    ls.sort();

    let mut l = 0;
    let mut r = 2;

    let mut total = 0;
    while l + 1 < r && r < n {
        for m in (l+1)..r {
            if ls[r] < ls[l] + ls[m] {
                total += r - m;
                break;
            }

        }
        r += 1;
        if r == n {
            l += 1;
            r = l + 2;
        }
    }

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

fn main() {
    solve();
}