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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args : Array<String>) {
    val scanner = Scanner(System.`in`)
    val n = scanner.nextInt()
    val array = IntArray(n)
    for (i in 0..n-1) {
        array[i] = scanner.nextInt()
    }
    val sortedArray = array.sortedArrayDescending()

    var sum = 0
    var preMax = 0
    loop1@for (i in 0..n-3) {
        val max = sortedArray[i]
        val pm = preMax
        preMax = max
        if (pm == max) continue@loop1
        val half = max / 2
        var preMiddle = 0

        loop2@for (j in i+1..n-2) {
            val middle = sortedArray[j]
            val pm = preMiddle
            preMiddle = middle
            if (pm == middle) continue@loop2
            if (middle <= half) break@loop2
            var preMin = 0

            loop3@for (k in j+1..n-1) {
                val min = sortedArray[k]
                val pm = preMin
                preMiddle = min
                if (pm == min) continue@loop3

                if (max < middle + min) {
                    sum += 1
                }
            }
        }
    }

    println(sum)
}