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++ (GCC 9.2.1) (judge verdict: Accepted)

// doot diddly donger cuckerino Hahahahahah


#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
using namespace __gnu_cxx;
using namespace std;

using ll = long long;
using ld = long double;
using pll = pair<ll, ll>;
using vi = vector<int>;

template <typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

const ld PI = acos(-1);

template <typename T> inline T mini(T& x, T y) { return x = min(x, y); }
template <typename T> inline T maxi(T& x, T y) { return x = max(x, y); }
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
template <typename T, typename U>
std::ostream& operator<<(std::ostream& stream, const pair<T, U>& p) {
  return stream << p.first << " " << p.second;
}
template <typename T>
std::ostream& operator<<(std::ostream& out, const vector<T>& vec) {
  for (const T& x: vec) out << x << ' ';
  return out;
}
template <typename T>
std::istream& operator>>(std::istream& in, vector<T>& vec) {
  for(auto& x: vec) in >> x;
  return in;
}
// @formatter:on


class DTriangles {
public:
  
  void solve(std::istream& cin, std::ostream& cout) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
    }
    sort(a.begin(), a.end());
    int ans = 0;
    for (int i = 0; i < n - 2; ++i) {
      int k = i + 2;
      for (int j = i + 1; j < n; ++j) {
        while (k < n && a[i] + a[j] > a[k]) ++k;
        if (k > j) {
          ans += k - j - 1;
        }
      }
    }
    cout << ans << '\n';
  }
};

int32_t main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
	DTriangles solver;
	std::istream& in(std::cin);
	std::ostream& out(std::cout);
	solver.solve(in, out);
	return 0;
}