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 <iostream>
//#include <iomanip>
//#include <string>
#include <vector>
#include <algorithm>
//#include <utility>
//#include <set>
//#include <map>
//#include <queue>
//#include <deque>
//#include <bitset>
//#include <math.h>
using namespace std ;
using ll = long long ;
//using ld = long double ;
using vll = vector<ll> ;
//using vvll = vector<vll> ;
//using vc = vector<char> ;
//using vvc = vector<vc> ;
//using vb = vector<bool> ;
//using vvb = vector<vb> ;
//using pll = pair<ll,ll> ;
#define all(v) v.begin(),v.end()
//ll mod = 1000000007 ;
//long double pie = acos(-1) ;
//ll INF = 100000000 ;

//void yorn(bool a){if(a) cout << "Yes" << endl ; else cout << "No" << endl ;}
//void YorN(bool a){if(a) cout << "YES" << endl ; else cout << "NO" << endl ;}
//ll gcd(long long a,long long b){if(b==0) return a ; return gcd(b,a%b) ;}
//ll lcm(long long a,long long b){return a/gcd(a,b)*b ;}
//ll sa(long long a,long long b){if(a>b) return a-b ; return b-a ;}
//void mysort(vector<long long> &a){sort(a.begin(),a.end()) ;}
//void myrev(vector<long long> &a){reverse(a.begin(),a.end()) ;}
//void fix_cout(){cout << fixed << setprecision(20) ;}

int main(){
	ll n ; cin >> n ;
	vll a(n) ;
	for(auto &i:a) cin >> i ;
	sort(all(a)) ;
	reverse(all(a)) ;
	ll ans = 0 ;
	for(int i=0;i<n-2;i++){
		for(int j=i+1;j<n-1;j++){
			if(a.at(j)*2<=a.at(i)) break ;
			ll l,r,m ;
			l = j ;
			r = n ;
			while(l+1<r){
				m = (l+r)/2 ;
				if(a.at(m)+a.at(j)>a.at(i)) l = m ;
				else r = m ;
			}
			ans += l-j ;
		}
	}
	cout << ans << endl ;
}