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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace::std;

#define all(x) (x).begin(), (x).end()

typedef long long ll;
typedef array<int, 3> tri;
typedef long double ld;

template <class T> istream& operator>>(istream& I, vector<T>& v) {for (T &e: v) I >> e; return I;}
template <class T> ostream& operator<<(ostream &O, const vector<T>& v) {for (const T &e: v) O << e << ' '; return O;}



void _main() {
        int n; cin >> n;
        vector<int> sticks(n); cin >> sticks;
        sort(all(sticks));
        //a, b, c [a <= b <= c]
        //a + b > c


        //a + b
        int ans = 0;
        for (int i = 0; i < n; i++) {
                for (int j = i + 1; j < n; j++) {
                        int a = sticks[i];
                        int b = sticks[j];
                        //c < a + b
                        //j ...
                        int l = j + 1, r = n - 1, where = j;
                        while (l <= r) {
                                int mid = (l + r) / 2;
                                int c = sticks[mid];
                                if (c < a + b)  {
                                        where = mid;
                                        l = mid + 1;
                                } else {
                                        r = mid - 1;
                                }
                        }
                        ans += where - j;
                }
        }

        cout << ans;
}



signed main() {
        ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
//        freopen("input.txt", "r", stdin);
        int _t = 1;
//        cin >> _t;
        while (_t--) _main();
        return 0;
}