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 PHP (5.6.30) (judge verdict: Wrong Answer)

<?php
$N = fgets(STDIN);
$L = fgets(STDIN);
$arrayL = explode(" ", $L);
sort($arrayL);

$result = 0;
// 1辺目
for ($i = 0; $i < $N - 2; $i ++) {
    // 2辺目
    for ($j = $i + 1; $j < $N - 1; $j ++) {
        // 3辺目
        // 許容最大値
        $longest = $arrayL[$i] + $arrayL[$j] - 1;
        // 次の値が許容最大値を超えていたら三角形は作れない
        if ($arrayL[$j + 1] > $longest) {
            continue;
        }
        $l = $j + 1;
        $r = $N - 1;
        $index = - 1;
        // 許容最大値から見つかるまで1ずつ引いていき探す(事前に見つかるか調べているので必ず見つかる)
        while ($index == - 1) {
            $index = binSearchOrigin($longest, $arrayL, $l, $r);
            $longest --;
        }
        // 数の重複対応、一番右にする
        if ($index != $N - 1) {
            while ($arrayL[$index] === $arrayL[$index + 1]) {
                $index ++;
                if ($index == $N - 1) {
                    break;
                }
            }
        }
        $result += $index - $j;
    }
}
echo $result;

// 二分探索_基本形
function binSearchOrigin($target, $array, $l, $r)
{
    if ($l > $r) {
        return - 1;
    }
    // 小数点切捨て
    $m = (int) (($l + $r) / 2);
    if ($target < $array[$m]) {
        return binSearchOrigin($target, $array, $l, $m - 1);
    } elseif ($target > $array[$m]) {
        return binSearchOrigin($target, $array, $m + 1, $r);
    } else {
        return $m;
    }
}