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)

#![allow(non_snake_case)]
#![allow(dead_code)]
#![allow(unused_macros)]
#![allow(unused_imports)]

use std::cmp::*;
use std::collections::*;

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};

    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
    ($iter:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    }
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };

    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };

    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };

    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };

    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

macro_rules! chmax {
    ($x:expr, $e:expr) => {
        if $x < $e {
            $x = $e;
        }
    };
}
macro_rules! chmin {
    ($x:expr, $e:expr) => {
        if $x > $e {
            $x = $e;
        }
    };
}

const INF: i64 = 1 << 60;

fn can_make(a: usize, b: usize, c: usize) -> bool {
    a < b + c && b < c + a && c < a + b
}

fn binary_search<F>(begin: usize, end: usize, solve: F) -> usize
    where F: Fn(usize) -> bool {
    let mut ok = begin - 1;
    let mut ng = end + 1;
    while ng - ok > 1 {
        let mid = (ng + ok) / 2;
        if solve(mid) {
            ok = mid;
        } else {
            ng = mid;
        }
    }
    ok
}

fn main() {
    input!{
        N: usize,
        mut L: [usize; N],
    }
    L.sort_by(|a, b| b.cmp(a));
    let mut ans = 0;
    for i in 0..(N-2) {
        let a = L[i];
        for j in (i+1)..(N-1) {
            let b = L[j];
            let cmax = binary_search(j+1, N-1, |k| {
                a - b < L[k] && L[k] < a + b
            });
            ans += cmax - j;
        }
    }
    println!("{}", ans);
}