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: Time Limit Exceeded)

fn read<T: std::str::FromStr>() -> Vec<T>
where
    <T as std::str::FromStr>::Err: std::fmt::Debug,
{
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).ok();
    buf.trim().split(' ').map(|x| x.parse().unwrap()).collect()
}

fn main() {
    let n = read::<usize>()[0];
    let mut l: Vec<i32> = read();
    l.sort_by_key(|x| -x);

    let mut ans = 0;

    for i in 0..(n - 2) as usize {
        for j in (i + 1)..(n - 1) as usize {
            let min = l[i] - l[j] + 1;
            if min >= l[j] {
                break;
            }
            let mut left: usize = std::cmp::min(j + 1, n - 1);
            let mut right: usize = n - 1;
            loop {
                if left == right {
                    break;
                }
                if left + 1 == right && l[left] >= min && l[right] >= min {
                    left = right;
                    break;
                }
                let next = (left + right) / 2;
                if l[next] > min {
                    left = next;
                } else {
                    right = next;
                }
            }
            if l[left] < min {
                left = j;
            }
            let target = l[left];
            while left + 1 < n && l[left + 1] == target {
                left += 1;
            }

            ans += left - j;
        }
    }

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