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)

#include<iostream>
#include<algorithm> 

using namespace std;

#define rep(i, n) for(int i = 0; i < n; i++)


int main(){
    int n;
    cin >> n;
    int l[n];
    rep(i, n) cin >> l[i];
    int num_sm = ((n - 1) * (n - 2)) / 2;
    pair< int, pair<int, int> > sm[num_sm];
    int num = 0;
    sort(l, l + n);
    for(int i = 0; i < n - 2; i++){
        for(int j =  i + 1; j < n - 1; j++){
            sm[num].first = l[i] + l[j];
            sm[num].second.first = l[i];
            sm[num].second.second = l[j];
            num++;
        }
    }
    // rep(i, num_sm) cout << sm[i].first << " " << sm[i].second.first << " " << sm[i].second.second << endl;
    
    int ans = 0;
    for(int i = 0; i < num_sm; i++){
        if (sm[i].second.first != sm[i].second.second) {
            int mx = lower_bound(l, l + n, sm[i].first) - l - 1;
            int mn = lower_bound(l, l + n, sm[i].second.second) - l;
            ans += mx - mn;
            // cout << mx << " " << mn << endl;
            // cout << ans << endl;
        }else{
            int mx = lower_bound(l, l + n, sm[i].first) - l - 1;
            int mn = lower_bound(l, l + n, sm[i].second.second) - l;
            ans += mx - mn - 1;
            // cout << mx << " " << mn << endl;
            // cout << ans << endl;  
        }
        

    }
    cout << ans << endl;   
}