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.Text;
using System.Collections.Generic;
using System.Linq;
using E = System.Linq.Enumerable;
 
class Program
 {
  
        static int[] L;
        
        static void Main(string[] args)
        {
          //入力を受け取る
          int N = int.Parse(Console.ReadLine());
          L = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
          L = L.OrderBy(a => -a).ToArray();
          long sum = 0;
          for(int i = 0 ;i < N ;i++){
            for(int j = i+1 ; j < N ;j++){
              //二分探索で何個三角形を作れるかを求める
              if(j+1 < N){
                sum += BinarySearchTriangle(i,j,j+1);
              }
            }
          }
          Console.WriteLine(sum);
        }
  
        static int BinarySearchTriangle(int i,int j,int k){
          int first = L[i];
          int second = L[j];
          //Console.WriteLine("first:"+first+"second:"+second+"third:"+L[k]);
          //二分探索の初期設定をする
          int head = k;
          int tail = L.Length-1;
          int mid = (head+tail)/2;
          if(mid != head){
            while(mid != head){
              if(first < second + L[mid]){
                head = mid;
                mid = (head+tail)/2;
              }else{
                tail = mid;
                mid = (head+tail)/2;
              }
            }
          }else if(first < second + L[mid]){
            return 1;
          }else{
            return 0;
          }
          //Console.WriteLine(head-j+1);
          return head-j+1;
        }
  
 }