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: Accepted)

#include <bits/stdc++.h>

#define taskname ""
#define pb push_back
#define eb emplace_back
#define fi first
#define se second
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define for0(i, n) for (int i = 0; i < (int)(n); ++i)
#define for1(i, n) for (int i = 1; i <= (int)(n); ++i)
#define ford(i, n) for (int i = (int)(n) - 1; i >= 0; --i)
#define fore(i, a, b) for (int i = (int)(a); i <= (int)(b); ++i)

using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex <ld> cd;
typedef vector <cd> vcd;
typedef vector <int> vi;

template<class T> using v2d = vector <vector <T> >;
template<class T> bool uin(T &a, T b)
{
    return a > b ? (a = b, true) : false;
}
template<class T> bool uax(T &a, T b)
{
    return a < b ? (a = b, true) : false;
}

mt19937 rng(chrono::system_clock::now().time_since_epoch().count());

const int maxN = 2e3 + 1;

int n;
int a[maxN];

ll res = 0;

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cin >> n;
    for1(i, n)
    {
        cin >> a[i];
    }
    sort(a + 1, a + n + 1);
    for (int i = n; i > 1; --i)
    {
        int l = 1, r = i - 1;
        while (l < r)
        {
            if (a[l] + a[r] > a[i])
            {
                res = res + r - l;
                --r;
            }
            else
            {
                ++l;
            }
        }
    }
    cout << res;
    return 0;
}