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 "bits/stdc++.h"
using namespace std;
#define INF 1000000000
#define rep(i, n) for(int i = 0; i < n; i++)
#define repr(i, n) for(int i = n; i >= 0; i--)
#define FOR(i, m, n) for(int i = m; i < n; i++)
#define ALL(x) (x).begin(), (x).end()  //昇順
#define RALL(x) (x).rbegin(), (x).rend() // 降順
#define SZ(x) ((int)(x).size)
const long long mod = 1e9+7;
typedef priority_queue<int> PQ;
typedef vector<long long> Vl;
typedef vector<int> VI;  // VI a(n);
typedef vector<string> VS;  
typedef vector<VS> VSS;  
typedef vector<VI> VVI;   // VVI a(n,vector<int>(m)) n * m 
typedef vector<Vl> VVl;
typedef pair<int, int> PII;
typedef map<int, int> MP;
typedef vector<pair<int, int> > PS;
typedef long long ll;  //ll とdoubleは違う
// cout << << endl;
// cin >> ;

int main(){
    ll x=1,g=0,n,h,w,k,m,ans=0,ptr=0;
    string s,s1,s2;
    cin >> n;
    VI l(n);
    rep(i,n){
        cin >> l[i];
	}
    sort(ALL(l));

    for(int i=n-1; i>=0; i--){
        for(int j=0; j<i;j++){
            for(int k=j+1; k<i;k++)
                if(l[i] < l[j]+l[k])
                    ans++;
        }
    }
    /*for(int i=0; i<n-2; i++){
        if(l[i+2] < l[i]+l[i+1] && l[i+1] < l[i+2]+l[i] && l[i] < l[i+1]+l[i+2] )
            ans++;
    }*/
    cout << ans << endl;
    return 0;
}