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: Time Limit Exceeded)

#include <iostream>
#include <algorithm>
#include <functional>
#include <vector>
#include <string>
#include <array>
#include <cmath>
using namespace std;
using ll = long long;

int main() {
	
	int n;
	cin >> n;
	vector<int> l(n);
	for (int i = 0; i < n; ++i) cin >> l.at(i);

	int res = 0;
	for (int i = 0; i < n; ++i)
	{
		for (int j = 0; j < n; ++j)
		{
			if (i == j) continue;
			for (int k = 0; k < n; ++k)
			{
				if (k == i || k == j) continue;
				if (l.at(i) < l.at(j) + l.at(k) && l.at(j) < l.at(i) + l.at(k) && l.at(k) < l.at(i) + l.at(j))
				{
					res++;
				}
			}
		}
	}

	if (res != 0) cout << res / 6 << endl;
	else cout << res << endl;

	return 0;
}