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)

// D - Triangles
// 累積和による方法
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vl = vector<ll>;
#define rp(i,s,e) for(int i=(s);i<(e);++i)

ll solve(int N, vi &L){
	vl csum(2000+1); //[length]<count>: 累積和: cumulative sum
	rp(i, 0, N) csum[L[i]]++;
	rp(i, 1, csum.size()) csum[i] += csum[i-1]; // その長さまでの棒の個数
	ll ans = 0;
	rp(ia, 0, N-1){
		rp(ib, ia+1, N){
			int a = L[ia], b = L[ib];
			int minL = abs(a - b), maxL = a + b; // |a-b| < c < a+b : cの範囲
			ans += csum[maxL-1] - csum[minL]; // maxL-1 に注意！cの範囲に(等号はない)
			if(minL<a && a<maxL) ans--; // cの範囲にaがあれば一本不要
			if(minL<b && b<maxL) ans--; // cの範囲にbがあれば一本不要
		}
	}
	return ans/3; //3本で△1つ
}

int main(){
	int N; cin>>N;
	vi L(N); rp(i, 0, N) cin>>L[i];
	cout<< solve(N, L) <<endl;
}