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: Wrong Answer)

use std::io;
use std::str::FromStr;

fn input<T:FromStr>() -> T {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.trim().parse().ok().unwrap()
}
fn input_i32_vec() -> Vec<i32> {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    line.split_whitespace().map(|s| s.parse().unwrap()).collect()
}

fn bs(left: usize, right: usize, l_vec: &Vec<i32>, target: i32) -> usize {
    // println!("l: {}, r: {}, target: {}", left, right, target);
    let mid = (left + right) / 2;
    if right - left <= 1 {
        // println!("{}", left);
        return left;
    }
    if l_vec[mid] >= target {
        bs(left, mid, l_vec, target)        
    } else {
        bs(mid, right, l_vec, target)
    }
}

fn main() {
    let n: i32 = input();
    let mut l_vec = input_i32_vec();
    l_vec.sort();
    println!("{:?}", l_vec);
    let mut sum = 0;
    for i in 0..(l_vec.len() - 2) {
        for j in (i+1)..(l_vec.len()-1) {
            // println!("{}, {}", i, j);
            let mut count = 0;
            if l_vec[i] + l_vec[j] <= l_vec[j+1] {
                continue;
            } else {
                sum += bs(j+1, l_vec.len(), &l_vec, l_vec[i] + l_vec[j]) - j;
            }
        }
    }
    println!("{}", sum);
}