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(dead_code)]
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()
}
#[allow(dead_code)]
fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}
#[allow(dead_code)]
fn read_vec2<T: std::str::FromStr>(n: u32) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}
#[allow(dead_code)]
fn yn(result: bool) {
    if result {
        println!("Yes");
    } else {
        println!("No");
    }
}

fn main() {
    let n = read::<usize>();
    let mut ls = read_vec::<u32>();
    ls.sort();

    //println!("sorted {:?}", ls);
    let mut count = 0;

    let lsmax = ls[n-1];

    /*
    let mut subcount = 0;
    for a in 0..n-2 {
        for b in a+1..n-1 {
            for c in b+1..n {
                if ls[a] < ls[b] + ls[c]
                    && ls[b] < ls[c] + ls[a]
                    && ls[c] < ls[a] + ls[b] {
                    //println!("a, b, c = ({}, {}, {})", a, b, c);
                    subcount += 1;
                }
            }
        }
    }
    */
    //println!("subcount: {}", subcount);

    for a in 0..n-2 {
        if ls[a] > lsmax / 2 + 1{
            let x = (n - a - 1) * (n - a - 2) / 2;
            count += x;
            //println!("a={}, x={}", a, x);
            continue;
        }
        for b in a+1..n-1 {
            let ab = ls[a] + ls[b];
            if ab > lsmax {
                let y = (n - b - 1) * (n - b) / 2;
                count += y;
                //println!("a={}, b={}, y={}", a, b, y);
                break;
            }

            for c in b+1..n {
                if ls[c] < ab {
                    count += 1;
                    //println!("a={}, b={}, c={}, +1", a, b, c);
                } else {
                    //println!("a={}, b={}, c={}, skip-c", a, b, c);
                    break;
                }
            }
        }
    }
    println!("{}", count);
}