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 <algorithm>
#include <bitset>
#include <iostream>
#include <vector>

#define rep(inc, bgn, end) for (int inc = bgn; inc < end; ++inc)
#define repe(inc, bgn, end) for (int inc = bgn; inc <= end; ++inc)

using ll = long long;
using namespace std;
int nextComb(int sub) {
    int x = sub & -sub;
    int y = sub + x;
    return (((sub & ~y) / x) >> 1) | y;
}

int main() {
    int n;
    cin >> n;
    vector<int> arrow(n, 0);
    rep(i, 0, n) {
        cin >> arrow[i];
        cout << i << endl;
    }

    int ans = 0;
    // 選び方全列挙
    int slctCnt = 3;
    for (int bit = (1 << slctCnt) - 1; bit < (1 << n); bit = nextComb(bit)) {
        // 選ばれたもののインデックス
        vector<int> idx;
        rep(i, 0, n) {
            if (bit & (1 << i)) {
                idx.push_back(i);
            }
            if (idx.size() == slctCnt) {
                break;
            }
        }

        if (arrow[idx[0]] <= arrow[idx[1]] + arrow[idx[2]] &&
            arrow[idx[1]] <= arrow[idx[2]] + arrow[idx[0]] &&
            arrow[idx[2]] <= arrow[idx[0]] + arrow[idx[1]]) {
            ans++;
        }
    }

    cout << ans << endl;

    return 0;
}