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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <algorithm>
#include <functional>
#include <string>
#include <vector>
#include <utility>
#include <climits>
#include <iomanip>
using namespace std;
#define MAXSIZE 100005
#define MAXL 100
#define MOD 998244353
typedef long long int lli;

int bs(vector<int> &len, int f, int s){
    int l = 0, r = s;
    int mid = 0;
    int hit = -1;
    do{
        if(mid < s && len[f] + len[mid] > len[s] && len[mid] + len[s] > len[f] && len[s] + len[f] > len[mid]){
            hit = mid;
            r = mid;
            // cout << "hit!" << endl;
            // cout << "s = " << s << ", hit = " << hit << ", f = " << f << endl;
        }else{
            l = mid;
        }
        // cout << "l = " << l << ", r = " << r << endl;
        mid = r - (r - l) / 2;
    }while(r - l >= 2);

    // cout << "s = " << s << ", hit = " << hit << ", f = " << f << endl << endl;
    return (hit == -1) ? 0 : s - hit;
}

int main(){
    int n;
    cin >> n;
    vector<int> len(n);
    for(int i = 0; i < n; ++i){
        cin >> len[i];
    }

    sort(len.begin(), len.end());

    lli ans = 0;
    for(int f = n - 1; f >= 2; --f){
        for(int s = f - 1; s >= 1; --s){
            ans += bs(len, f, s);
        }
    }

    cout << ans << endl;
    return 0;
}