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)

#![allow(non_snake_case)]
#![allow(unused_imports)]
use proconio::input;
use std::cmp::max;
use std::cmp::min;
use std::collections::HashMap;
use std::collections::VecDeque;

fn main() {
    input! {
        N: usize,
        tmpLs: [i64; N],
    }

    let mut Ls: Vec<i64> = tmpLs.into_iter().collect();

    // ソートはタダ。
    Ls.sort();

    let mut ans = 0i64;

    for i in 0..N {
        let A = Ls[i];

        for j in i + 1..N {
            let B = Ls[j];

            // 3本目は、 j < k < N であるkで、
            // L[k] < A + Bを満たすものの中から選ぶ

            // L[k] < A + Bを満たすkは二分探索で求まりそう

            let sum_l = A + B;

            let mut ok = j + 1;
            if ok == N - 1 && Ls[ok] < A + B {
                ans += 1;
                continue;
            }

            if ok > N - 1 || Ls[ok] >= A + B {
                continue;
            }

            let mut ng = N;

            while ng - ok > 1 {
                let mid = (ok + ng) / 2;
                if Ls[mid] < sum_l {
                    ok = mid;
                } else {
                    ng = mid;
                }
            }

            ans += (ok - j) as i64;
        }
    }

    println!("{}", ans);
}