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::{self, BufRead};

pub fn lower_bound<T: Ord>(vec: &[T], val: T) -> usize {
  let mut lo = 0;
  let mut hi = vec.len();
  while hi - lo > 0 {
    let p = (hi + lo) / 2;
    if vec[p] < val {
      lo = p + 1;
    } else {
      hi = p;
    }
  }
  return lo;
}

pub fn upper_bound<T: Ord>(vec: &[T], val: T) -> usize {
  let mut lo = 0;
  let mut hi = vec.len();
  while hi - lo > 0 {
    let p = (hi + lo) / 2;
    if vec[p] > val {
      hi = p;
    } else {
      lo = p + 1;
    }
  }
  return lo;
}

fn main() {
  let stdin = io::stdin();
  let mut lines = stdin.lock().lines();
  let n: usize = lines.next().unwrap().unwrap().parse().unwrap();
  let mut ls: Vec<usize> = lines.next().unwrap().unwrap().split(' ').map(|x| x.parse().unwrap()).collect();
  ls.sort();
  let mut count = 0;
  for i in 0..(n-2) {
    for j in (i+1)..(n-1) {
      let x = ls[i];
      let y = ls[j];
      let min = y - x + 1;
      let max = x + y - 1;
      let lo = lower_bound(&ls[(j+1)..], min);
      let hi = upper_bound(&ls[(j+1)..], max);
      count += hi - lo;
    }
  }
  println!("{}", count);
}