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)

/*
  author: ruruvuvu
  GitHub account: colon-ku

  created at: 2019-12-13 16:12:41
*/

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

#define all(x) (x).begin(),(x).end()
#define rep(x, y) for (int x = 0; x < y; x++)
#define MOD 1000000007

typedef long long LL;
typedef long double LD;

bool compare_by_b(pair<int, int> a, pair<int, int> b)
{
    if (a.second != b.second) {
        return a.second < b.second;
    } else {
        return a.first < b.first;
    }
}

ostringstream oss_global;
string s_global = oss_global.str();

int search(vector<int> v, int key)
{
    // key より大きい要素の中で一番小さいものを選ぶ
    int n, left, right, mid;
    n = v.size(); left = 0; right = n;

    while (left < right) {
        mid = (left + right) / 2;
        if (v[mid] == key) {
            return mid + 1;
        } else if (v[mid] > key) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }

    return -1;
}

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

    sort(all(L));
    
    int ans = 0;

    if (L.size() == 3) {
        if (L[2] < L[0] + L[1]) ans = 1;
    }

    for (int i = N-1; i >= 2; i--) {
        for (int j = i-1; j >= 1; j--) {
            int index = search(L, L[i] - L[j]);
            if (index > 0 && index < j) {
                ans += (j - index);
            }
        }
    }

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