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 (1.42.0) (judge verdict: Accepted)

use proconio::*;
use std::cmp::*;
use std::vec::Vec;
use std::cmp::Ordering;


/// Equivalent to std::lowerbound and std::upperbound in c++
pub trait BinarySearch<T> {
    fn lower_bound(&self, x: &T) -> usize;
    fn upper_bound(&self, x: &T) -> usize;
}

impl<T: Ord> BinarySearch<T> for [T] {
    fn lower_bound(&self, x: &T) -> usize {
        let mut low = 0;
        let mut high = self.len();

        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(x) {
                Ordering::Less => {
                    low = mid + 1;
                }
                Ordering::Equal | Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }

    fn upper_bound(&self, x: &T) -> usize {
        let mut low = 0;
        let mut high = self.len();

        while low != high {
            let mid = (low + high) / 2;
            match self[mid].cmp(x) {
                Ordering::Less | Ordering::Equal => {
                    low = mid + 1;
                }
                Ordering::Greater => {
                    high = mid;
                }
            }
        }
        low
    }
}

fn main() {
    input! {
        n: usize,
        mut L: [i32; n],
    }
    L.sort();
    let mut res = 0;
    for a in 0..n{
        for b in a + 1..n{
            let ma = L[a] + L[b];
            let mi = L[b] - L[a];
            let left = L.upper_bound(&mi);
            let right = L.lower_bound(&ma);
            let mut cnt = right - left;
            if mi < L[a] && L[a] < ma {
                cnt -= 1;
            }
            if mi < L[b] && L[b] < ma {
                cnt -= 1;
            }
            res += cnt;
        }
    }
    println!("{}", res / 3);
}