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

#define pb push_back
#define F first
#define S second
#define fo(i,a,b)  for(int i = a; i < b; i++)

using ll = long long;
using ld = long double;
using ul = unsigned long long;
using ii = pair<int, int>;
#define Mosta     ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ReadFile            freopen("input.in","r",stdin)
#define WriteFile           freopen("output.txt","w",stdout)
const ll mod7 = 1000000007,mod = 1000000000,OO = 0x3f3f3f3f;
const ld EPS = 1e-7;
double Pi=acos(-1);
const ld rad=Pi/180.0;
ll power(ll x, ll y, ll p)
{
    ll res = 1;
    while (y > 0)
    {
        if (y & 1)
            res = (res*x) % p;
        y = y>>1;
        x = (x*x) % p;
    }
    return res;
}
ll nCr (ll n,ll m)
{
    return(m == 0)?1:n*nCr(n-1,m-1)/m;
}
/*****************************************************/

const int N = (int) 2e3 + 4;

int arr[N];

int n , sum , ans = 0 , i , j;

bool valid(int mid){
    if(sum > mid)
        return true;
    return false;
}

void bs(){
    int l = 1 , r = n;
    while(l <= r){
        int m = l + (r - l) / 2;
        if(valid(arr[m]) && m != i && m != j && m > i)
            ans++ , l = m + 1;
        else r = m - 1;
    }
}

int main()
{
    cin >> n;

    for(int i = 1; i <= n; ++i)
        cin >> arr[i];

    sort(arr , arr + n);

    for(i = 1; i <= n; ++i){
        for(j = i+1; j <= n; ++j){
            sum = arr[i] + arr[j];
            bs();
        }
    }

    cout << ans << "\n";

    return 0;
}