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)

// Undone
#include <bits/stdc++.h>
#define int long long
#define double long double
#define INF 1e18
#define MOD 1000000007
#define FAST_IO {ios::sync_with_stdio(0);cin.tie(0);}
using namespace std;
signed main() {
  int N; cin >> N;
  vector<int> L(N);
  for (int i = 0; i < N; i++) cin >> L[i];
  sort(L.begin(), L.end());
  
  set< pair<int, pair<int,int> > > tris;
  for (int i = 0; i < N; i++) {
    for (int j = i+1; j < N; j++) {
      int a = L[j];
      int b = L[i];
      auto itr = upper_bound(L.begin(), L.end(), a-b);
      if (itr == L.end()) break;
      while (itr != L.end()) {
        int c = *itr;
        if (a-b < c && c < a+b && c != a && c != b) {
          int x = min(c, min(a, b));
          int z = max(c, max(a, b));
          int y;
          if ((x == b && z == c) || (x == c && z == b)) {
            y = a;
          } else if ((x == a && z == c) || (x == c && z == a)) {
            y = b;
          } else {
            y = c;
          }
          tris.insert(make_pair(x, make_pair(y, z)));
        }
        itr++;
      }
    }
  }
  cout << tris.size() << endl;
  /*
  for (auto itr = tris.begin(); itr != tris.end(); itr++) {
    pair< int, pair<int, int> > p = *itr;
    cout << p.first << "," << p.second.first << "," << p.second.second << endl;
  }
  */
}