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: Wrong Answer)

  //shan61916
#include <bits/stdc++.h>
  using namespace std;
typedef   long long ll;
typedef   unsigned long long ull ;
typedef   double dll ;

#define   mp make_pair
#define   pb push_back
#define   ff first 
#define   ss second
#define   IOS ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define   endl "\n" 
#define   pll pair<ll, ll>
#define   all(x) x.begin(), x.end()
#define   vll vector<ll> 


int main(){
 IOS
#ifdef SHAN
    freopen("input.txt" , "r" , stdin);  
#endif
  ll n;
  cin >> n;
  vector<ll> a(n);
  for(auto &x: a) cin >> x;
  sort(all(a));
  ll ans = 0;
  for(ll i = 0; i < n; i++) {
  	for(ll j = i+1; j < n; j++) {
  		ll x = a[i];
  		ll y = a[j];
  	    ll from = (x + y)/2; 
  	    ll to = (x+y-1);
  	    ll idx1 = upper_bound(a.begin()+j+1, a.end(), from) - a.begin();
  	    auto idx2 = lower_bound(a.begin()+j+1, a.end(), to);
  	    if(*idx2 != to) idx2--;
  	    ll id2 = idx2 - a.begin();
  	    if(idx1 <= id2) ans+= id2-idx1+1;
  }
}
  cout << ans << endl;
  return 0;
} //good night.