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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int main() {
	int N, counter = 0,flag=0;
	cin >> N;
	vector<int> Length(N);
	for (size_t i = 0; i < N; i++)
	{
		cin >> Length.at(i);
      Length.at(i)*=-1;
	}
	sort(Length.begin(), Length.end());
  		for (size_t i = 0; i < N; i++)
	{
      Length.at(i)*=-1;
	}
	for (size_t i = 0; i < N; i++)
	{
		flag = 0;
		for (size_t j = i+1; j < N; j++)
		{
			for (size_t k = j+1; k < N; k++)
			{
				if (Length.at(i)<(Length.at(j)+Length.at(k))&&abs(Length.at(j)-Length.at(k))<Length.at(i))
				{
					counter++;
				}
				else {
					flag = 1;
					break;
				}
				
			}
			if (flag == 1) {
              	flag=0;
				break;
			}
		}

	}
	cout << counter << endl;
}