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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using System.Text;


class Program
{
    static void Main(string[] args)
    {
        // 入力
        //var s = Console.ReadLine();
        var n = int.Parse(Console.ReadLine());
        int[] l = Console.ReadLine().Split().Select(int.Parse).ToArray();

        var dic = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
        {
            if (!dic.ContainsKey(l[i]))
            {
                dic.Add(l[i], 1);
            }
            else
            {
                dic[l[i]]++;
            }
        }

        Array.Sort(l);
        Array.Reverse(l);
        var ans = 0;
        for (int i = 0; i < n; i++)
        {
            for (int k = i + 1; k < n; k++)
            {
                //l[i]が一番目に長い
                //l[k]が二番目に長いとする
                if (l[i] < l[k]) continue;
                var z1 = l[i] + l[k];
                var z2 = Math.Max(l[i], l[k]) - Math.Min(l[k], l[i]);
                for (int t = z2 + 1; t < z1; t++)
                {
                    if (l[i] < t || l[k] < t) continue;                   
                    if (dic.ContainsKey(t))
                    {
                        if (t == l[i] || t == l[k])
                        {
                            ans += dic[t] - 1;
                        }
                        else
                        {
                            ans += dic[t];
                        }
                    }
                }
            }
        }

        // 出力
        Console.WriteLine(ans);
    }
}