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: Accepted)

#include <bits/stdc++.h>
#define int long long
#define mod (int)(1e9+7)
#define inf (int)(3e18)
#define rep(i,n) for(int i=0;i<n;i++)
#define REP(i,n) for(int i=1;i<n;i++)
#define P std::pair<int,int>
#define PiP std::pair<int,std::pair<int,int>>
#define all(v) v.begin(),v.end()
#define mkp std::make_pair
#define prique(T) std::priority_queue<T,vector<T>,greater<T>>
using namespace std;
template<class T> inline void chmax(T& a, T b) { a = std::max(a, b); }
template<class T> inline void chmin(T& a, T b) { a = std::min(a, b); }

bool prime(int x) {
	for (int i = 2; i * i <= x; i++) {
		if (x % i == 0)return false;
	}
	return x != 1;
}
int gcd(int x, int y) {
	if (y == 0)return x;
	return gcd(y, x % y);
}
int lcm(int x, int y) {
	return x / gcd(x, y) * y;
}
int kai(int x, int y) {
	int res = 1;
	for (int i = x - y + 1; i <= x; i++) {
		res *= i; res %= mod;
	}
	return res;
}
int mod_pow(int x, int y, int m) {
	int res = 1;
	while (y > 0) {
		if (y & 1) {
			res = res * x % m;
		}
		x = x * x % m;
		y >>= 1;
	}
	return res;
}
int comb(int x, int y) {
	if (y > x)return 0;
	return kai(x, y) * mod_pow(kai(y, y), mod - 2, mod) % mod;
}
int get_rand(int MIN, int MAX) {
	std::random_device rnd;
	std::mt19937_64 mt64(rnd());
	std::uniform_int_distribution<int>engine(MIN, MAX);
	return engine(mt64);
}
/*--------Library Zone!--------*/

int n, a[2005];
signed main() {
	cin >> n;
	rep(i, n)cin >> a[i];
	sort(a, a + n);
	int ans = 0;
	rep(i, n) {
		for (int j = i + 1; j < n; j++) {
			int maindex = upper_bound(a, a + n, max(a[i] - a[j], a[j] - a[i])) - a;
			if (maindex <= j)maindex = j + 1;
			int miindex = lower_bound(a, a + n, a[i] + a[j]) - a;
			ans += max(0ll, miindex - maindex);
		}
	}
	cout << ans << endl;
}