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 (N) = readLine()!!.split(" ").map { it.toInt() }
    val L =readLine()!!.split(" ").map { it.toInt() }.toIntArray()
    L.sort()
    var ans = 0
    for (a in 0 until N - 2){
        for (b in (a+1) until N - 1) {
            var c = bsearch(L, L[a] + L[b], b + 1, N-1)
            if (c == N) c = N -1
            // println("a=$a(${L[a]}) b=$b(${L[b]}) c=$c(${L[c]}) ${c - b}")
            ans += c - b
        }
    }
    println(ans)
}

// arrayの中でvの値を探し、そのIndexを返す
fun bsearch(array: IntArray, v: Int, min: Int, max: Int) : Int {
    if (min > max) return min
    val mid = min + (max - min) / 2
    return when {
        array[mid] > v -> bsearch(array, v, min, mid-1)
        array[mid] < v -> bsearch(array, v, mid + 1, max)
        else -> mid
    }
}