Problem:
Score : 100 points

Problem Statement

The window of Takahashi's room has a width of A. There are two curtains hung over the window, each of which has a horizontal length of B. (Vertically, the curtains are long enough to cover the whole window.)

We will close the window so as to minimize the total horizontal length of the uncovered part of the window.
Find the total horizontal length of the uncovered parts of the window then.

Constraints

1 \leq A \leq 100

1 \leq B \leq 100

A and B are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the total horizontal length of the uncovered parts of the window.

Sample Input 1

12 4

Sample Output 1

4

We have a window with a horizontal length of 12, and two curtains, each of length 4, that cover both ends of the window, for example. The uncovered part has a horizontal length of 4.

Sample Input 2

20 15

Sample Output 2

0

If the window is completely covered, print 0.

Sample Input 3

20 30

Sample Output 3

0

Each curtain may be longer than the window.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define mod 1000000007
#define all(v)  v.begin(),v.end()
#define pb push_back
#define size(v)  (int) v.size()
#define fast ios_base:: sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL)
ll power_mod(ll a,ll x){if(x==0) return 1;ll y=power_mod(a,x/2);ll ans=(y*y)%mod; if(x%2) ans=(ans*a)%mod;return ans;}
ll inv(ll a){return power_mod(a,mod-2);}    
ll power(ll a, ll x){ if(x==0) return 1; ll y= power(a,x/2); ll ans=(y*y); if(x%2) ans*=a; return ans;}

int main(){
  int a,b;
  cin>>a>>b;
  int c = a-2*b;
  cout<<max(0,c)<<endl;
  
}