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 <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#define all(x) (x).begin(),(x).end()
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define REP(i, m, n) for(int i = m; i < n; i++)
#define INF INT_MAX
#define MOD 1000000007
#define fcout cout << fixed << setprecision(15)
typedef long long ll;
typedef pair<int, int> P;
int gcd(int a,int b){return b?gcd(b,a%b):a;};
int lcm(int a,int b){return a * b / gcd(a, b);};
int mod(int a,int b){return (a+b-1)/b;};
int month[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
template<class T>inline bool chmax(T& a,T b){if(a < b){a=b;return true;}return false;};
template<class T>inline bool chmin(T& a,T b){if(a > b){a=b;return true;}return false;};

bool tri(int a, int b, int c) {
  if(abs(a - b) < c && c < a + b) return true;
  else return false;
}

struct san {
  int x, y, z;
};

int main() {
  cin.tie(0);
  cout.tie(0);
  ios::sync_with_stdio(false);

  int n;
  cin >> n;
  int l[n];
  rep(i, n) cin >> l[i];
  set< tuple<int, int, int> > res;
  tuple<int, int, int> t;
  sort(l, l + n);
  rep(i, n) {
    REP(j, i + 1, n) {
      REP(k, j + 1, n) {
        if(tri(l[i], l[j], l[k])) {
          t = make_tuple(l[i], l[j], l[k]);
          res.insert(t);
        }
      }
    }
  }
  cout << res.size() << endl;
  return 0;
}