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

namespace AtCoder.D
{
    public class Program
    {
        public static void Main() { var r = GetResult(); Debug.WriteLine(r); Console.Write(r); }

        private static object GetResult()
        {
            var N = ReadLong();
            var L = ReadLongs().OrderByDescending(x => x).ToArray();

            var count = 0L;
            for (int i = 0; i < N - 2; i++)
            {
                for (int j = i + 1; j < N - 1; j++)
                {
                    var k = L[i] - L[j];
                    var l = j + 1;
                    var r = N - 1;
                    while (true)
                    {
                        if (L[l] <= k)
                        {
                            break;
                        }
                        if (L[r] > k)
                        {
                            count += r - j;
                            break;
                        }
                        var mid = (l + r) / 2D;
                        var up = (int)Math.Ceiling(mid);
                        if (k >= L[up])
                        {
                            r = (int)Math.Floor(mid);
                        }
                        else
                        {
                            l = up;
                        }

                        if (r - l == 1 || r == l)
                        {
                            count += r - j;
                            break;
                        }
                    }
                }
            }

            return count;
        }

        #region Console

        public static string ReadText() { return Console.ReadLine(); }
        public static List<string> ReadTexts() { return Console.ReadLine().Split(' ').ToList(); }
        public static long ReadLong() { return long.Parse(Console.ReadLine()); }
        public static List<long> ReadLongs() { return Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToList(); }

        #endregion
    }
}