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<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<map>
#include<string.h>
#include<complex>
#include<math.h>
#include<queue>
#include <functional>
#include<time.h>
#include <stack>
#include<iomanip>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define ll long long
#define llint long long int
#define sort(s) sort(s.begin(),s.end())
#define reverse(v)  reverse(v.begin(), v.end());
#define Yes(ans) if(ans)cout<<"Yes"<<endl; else cout<<"No"<<endl;
#define YES(ans) if(ans)cout<<"YES"<<endl; else cout<<"NO"<<endl;
#define hei(a) vector<a>
#define whei(a) vector<vector<a>>
#define UF  UnionFind
constexpr auto mod = 1000000007;
//辞書順はnext_permutation( begin( v ), end( v ) );やで！


//2のn乗を求めるよ！！！
int ni(int n) {
	if (n == 0)return 1;
	int x = ni(n / 2);
	x *= x;
	if (n % 2 == 1)x *= 2;
	return x;
}

//フィボナッチ数列のx番目を求めるよ！
llint f(int x, vector<llint> &s) {
	if (x == 0)return 0;
	if (x == 1)return 1;
	if (s[x] != 0)return s[x];
	return  s[x] = f(x - 1, s) + f(x - 2, s);
	
}

//aとｂの最大公約数を求めるよ！
llint gcd(llint a, llint b) {
	if (a < b)swap(a, b);
	if (b == 0)return a;
	return gcd(b, a % b);
}



// a^n mod を計算する
long long modpow(long long a, long long n, long long mod) {
	long long res = 1;
	while (n > 0) {
		if (n & 1) res = res * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return res;
}

// mod. m での a の逆元 a^{-1} を計算するよ！
long long modinv(long long a, long long m) {
	long long b = m, u = 1, v = 0;
	while (b) {
		long long t = a / b;
		a -= t * b; swap(a, b);
		u -= t * v; swap(u, v);
	}
	u %= m;
	if (u < 0) u += m;
	return u;
}


//aCbをmod.mで割った余りを求める
ll int c(ll int a, ll int b, ll int m) {
	ll int ans = 1;
	for (ll int i = 0; i < b; i++) {
		ans *= a - i;
		ans %= m;
	}
	for (ll int i = 1; i <= b; i++) {
		ans *= modinv(i, m);
		ans %= m;
	}
	return ans;
}

//m進数のn桁を全列挙やで 
int dfs(int m, int n,llint k,hei(llint) a) {
	stack<string> st;
	st.push("");
	while (!st.empty()) {
		string now = st.top();
		st.pop();
		if (now.size() == n) {
			llint ans = 0;
				rep(i, 0, n) {
					if (now[i] == '1')ans += a[i];
				}
				if (ans == k) {
					cout << "Yes" << endl;
					return 0;
				}
	
		}
		else {
			for (int i = m - 1; i >= 0; i--) {
				string next = now + to_string(i);
				st.push(next);
			}
		}
	}
	return 1;

}
void press(vector<int>& v) { v.erase(unique((v).begin(), (v).end()), v.end()); }
void press(vector<char>& v) { v.erase(unique((v).begin(), (v).end()), v.end()); }

llint min(llint a, llint b) {
	if (a < b)return a;
	return b;
}
llint max(llint a, llint b) {
	if (a < b)swap(a, b);
	return a;
}
llint p(int n) {
	if(n==1)return 1;
	return n * p(n - 1);
}
struct UnionFind {
	vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

	UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
		for (int i = 0; i < N; i++) par[i] = i;
	}
	int size(int a) {
		return par[root(a)];
	}

	int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
		if (par[x] == x) return x;
		return par[x] = root(par[x]);
	}

	void unite(int x, int y) { // xとyの木を併合
		int rx = root(x); //xの根をrx
		int ry = root(y); //yの根をry
		if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
		par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
	}

	bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
		int rx = root(x);
		int ry = root(y);
		return rx == ry;
	}
};

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