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)

// Try AtCoder
// author: Leonardone @ NEETSDKASU

fn main() {
    let mut stdin = String::new();
    std::io::Read::read_to_string(&mut std::io::stdin(), &mut stdin).unwrap();
    let mut stdin = stdin.split_whitespace();
    let mut get = || stdin.next().unwrap();
    macro_rules! get {
        ($t:ty) => (get().parse::<$t>().unwrap());
        () => (get!(i64));
    }
    
    let n = get!(usize);
    let mut ss = vec![0; 1001];
    for _ in 0 .. n {
        let s = get!(usize);
        ss[s] += 1;
    }
    
    let mut sum = vec![0; 1002];
    for i in 0 .. 1001 {
        sum[i+1] = sum[i] + ss[i];
    }
    let mut ans = 0_i64;
    for x in 1 .. 1001 {
        if ss[x] == 0 {
            continue;
        }
        if ss[x] >= 2 {
            for y in 1 .. 1001 {
                if y != x && y < x + x {
                    ans += (ss[x] * (ss[x] - 1) / 2) * ss[y];
                }
            }
        }
        if ss[x] >= 3 {
            ans += ss[x] * (ss[x] - 1) * (ss[x] - 2) / 6;
        }
        for y in 1 .. x {
            if ss[y] == 0 {
                continue;
            }
            for z in 1 .. y {
                if y < x + z && x < y + z && z < x + y {
                    ans += ss[x]*ss[y]*ss[z];
                }
            }
        }
    }
    println!("{}", ans);
}