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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <climits>
#include <cmath>
#include <algorithm>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <deque>
#define INF_INT (INT_MAX / 2)
#define INF_LONG (LONG_MAX / 2)
//#define DEBUG true
#define DEBUG false
using namespace std;

const int MAX = 100001;
const int MOD = 1000000007;

typedef long long ll;
typedef pair< int, int > pii;
typedef pair< ll, ll > pll;
typedef pair< ll, int > pli;
typedef pair< int, ll > pil;

int ceil(int x, int y){
  return (x % y == 0) ? x / y : x / y + 1;
}

int gcd(int x, int y){
  return y ? gcd(y, x % y) : x;
}

int lcm(int x, int y){
  return x / gcd(x, y) * y;
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int n;
  cin >> n;
  int l[n];
  for(int i = 0; i < n; i++) cin >> l[i];

  sort(l, l + n);

  int sum = 0;
  for(int i = 0; i < n; i++){
    int a = l[i];
    for(int j = i + 1; j < n; j++){
      int b = l[j];
      auto t = lower_bound(l + j + 1, l + n, a + b);
      auto s = upper_bound(l + j + 1, l + n, b - a);
      sum += distance(s, t);
    }
  }

  cout << sum << endl;

  return 0;
}