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

class Program
{
    static void Main(string[] args) {
#if DEBUG
        //MyDebugger.MakeTestCase("../TextFile2.txt");
        MyDebugger.Test(process, "../TextFile1.txt");
#else
        object ret = process(new StreamReader(Console.OpenStandardInput()));
        Console.WriteLine(ret);
#endif
    }
    

    static object process(TextReader input) {
        int n = int.Parse(input.ReadLine());
        //int[] ll = ToIntArray(input.ReadLine(), n);
        
        int[] tb = new int[1001];

        var spl =input.ReadLine().Split(' ');
        for (int i = 0; i < n; i++) {
            var x = int.Parse(spl[i]);
            tb[x]++;
        }

        long count = 0;

        for(int i=1000; i>0; i--) {
            //if (tb[i] == 0) continue;
            count += Combination(tb[i],3);

            for(int j=i-1; j>0; j--) {
                if (i > j * 2) break;
                //if (tb[j] == 0) continue;

                count += tb[i] * Combination(tb[j],2);

                for(int k=j-1; k>0; k--) {
                    if (i > j + k) break;
                    //if (tb[k] == 0) continue;

                    count += tb[i] * tb[j] * tb[k];
                }
            }
        }

        return count;
    }

    static long Combination(long m, long n) {
        if (m < n) return 0;
        if (m == n || n == 0) return 1;

        long ans = 1;
        for (long i = 1; i <= n; i++) {
            ans = ans * (m - i + 1) / i;
        }
        return ans;
    }
}