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;
/*#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<typename T> using gpp_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename L> using gpp_map = tree<T, L, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T> using gpp_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;*/
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
#define FOR(i, begin, end) for(int i=(begin);i<(end);i++)
#define REP(i, n) FOR(i,0,n)
#define IFOR(i, begin, end) for(int i=(end)-1;i>=(begin);i--)
#define IREP(i, n) IFOR(i,0,n)
#define Sort(v) sort(v.begin(), v.end())
#define Reverse(v) reverse(v.begin(), v.end())
#define all(v) v.begin(),v.end()
#define SZ(v) ((int)v.size())
#define Lower_bound(v, x) distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) distance(v.begin(), upper_bound(v.begin(), v.end(), x))
#define Max(a, b) a = max(a, b)
#define Min(a, b) a = min(a, b)
#define bit(n) (1LL<<(n))
#define bit_exist(x, n) ((x >> n) & 1)
#define debug(x) cout << #x << "=" << x << endl;
#define vdebug(v) cout << #v << "=" << endl; REP(i_debug, v.size()){ cout << v[i_debug] << ","; } cout << endl;
#define mdebug(m) cout << #m << "=" << endl; REP(i_debug, m.size()){ REP(j_debug, m[i_debug].size()){ cout << m[i_debug][j_debug] << ","; } cout << endl;}
#define pb push_back
#define f first
#define s second
#define int long long
#define INF 1000000000000000000
template<typename T> istream &operator>>(istream &is, vector<T> &v){ for (auto &x : v) is >> x; return is; }
template<typename T> ostream &operator<<(ostream &os, vector<T> &v){ for(int i = 0; i < v.size(); i++) { cout << v[i]; if(i != v.size() - 1) cout << endl; }; return os; }
template<typename T> void Out(T x) { cout << x << endl; }
template<typename T1, typename T2> void Ans(bool f, T1 y, T2 n) { if(f) Out(y); else Out(n); }

using vec = vector<int>;
using mat = vector<vec>;
using Pii = pair<int, int>;
using PiP = pair<int, Pii>;
using PPi = pair<Pii, int>;
using bools = vector<bool>;
using pairs = vector<Pii>;

//int dx[4] = {1,0,-1,0};
//int dy[4] = {0,1,0,-1};
//char d[4] = {'D','R','U','L'};

const int mod = 1000000007;
//const int mod = 998244353;
//#define Add(x, y) x = (x + (y)) % mod
//#define Mult(x, y) x = (x * (y)) % mod



signed main(){

    int A, B; cin >> A >> B;
    Out(max(0LL, A - 2 * B));

    return 0;
}