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 <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <iomanip>
#include <set>
#include <queue>
#include <stack>
#include <utility>
using namespace std;
typedef long long int ll;

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define drep(i, n) for(int i = n - 1; i >= 0; i--)
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define yes cout << "yes" << endl
#define NO cout << "NO" << endl
#define No cout << "No" << endl
#define no cout << "no" << endl

const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;
const int mod = 1000000007;
#define PI 3.14159265359;

void P(int x) {cout << x << endl;}
void P(long x) {cout << x << endl;}
void P(double x) {cout << x << endl;}
void P(ll x) {cout << x << endl;}
void P(string x) {cout << x << endl;}
void P(char x) {cout << x << endl;}

ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}

int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  rep (i, n) {
    cin >> v[i];
  }
  sort(v.begin(), v.end());

  ll ans = 0;
  rep (i, n - 2) {
    for (int j = i + 1; j < n - 1; j++) {
      int sum = v[i] + v[j];
      for (int k = j + 1; k < n; k++) {
        if (v[k] < sum) {
          ans++;
          // cout << v[i] << " " << v[i + 1] << " " << v[j] << endl;
        } else {
          break;
        }
      }
    }
  }
  P(ans);

  return 0;
}