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)

use std::io::Read;

pub fn read_stdin() -> String {
    let mut s = String::new();
    ::std::io::stdin().read_to_string(&mut s).unwrap();
    s
}

pub trait Parse {
    fn to<T>(&mut self) -> T
    where
        T: std::str::FromStr;
}

impl<'a> Parse for std::str::SplitWhitespace<'a> {
    fn to<T>(&mut self) -> T
    where
        T: std::str::FromStr,
    {
        let word = self.next().unwrap();
        word.parse().ok().unwrap()
    }
}

fn comb2(n: usize) -> Vec<(usize, usize)> {
    let mut v = Vec::new();
    for i in 0..(n - 1) {
        for j in (i + 1)..n {
            v.push((i, j));
        }
    }
    v
}

trait BinarySearch<T> {
    fn front_position(&self, x: T) -> usize;
    fn back_position(&self, x: T) -> usize;
    fn bs_position(v: &[T], x: T, is_back: bool) -> usize;
}

impl<T> BinarySearch<T> for Vec<T>
where
    T: Ord,
{
    fn front_position(&self, x: T) -> usize {
        Self::bs_position(self, x, true)
    }

    fn back_position(&self, x: T) -> usize {
        Self::bs_position(self, x, false)
    }

    fn bs_position(v: &[T], x: T, is_front: bool) -> usize {
        let len = v.len();
        if len == 1 {
            return if x < v[0] || (is_front && x == v[0]) {
                0
            } else {
                1
            };
        }
        let i = len / 2;
        if x < v[i] || (is_front && x == v[i]) {
            Self::bs_position(&v[..i], x, is_front)
        } else {
            Self::bs_position(&v[i..], x, is_front) + i
        }
    }
}

fn main() {
    let source = read_stdin();
    let mut iter = source.split_whitespace();
    let n = iter.to();
    let mut v: Vec<i32> = (0..n).map(|_| iter.to()).collect();
    v.sort();
    let mut count = 0;
    for (i, j) in comb2(n) {
        let a = v[i];
        let b = v[j];
        let mut vv = v.clone();
        let rem_a = vv.front_position(a);
        vv.remove(rem_a);
        let rem_b = vv.front_position(b);
        vv.remove(rem_b);
        // a - b < c
        // b - a < c
        // c < a + b
        let max = std::cmp::max(a - b, b - a);
        let start = vv.back_position(max);
        let end = vv.front_position(a + b);
        count += end - start;
    }
    println!("{}", count / 3);
}