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 unsigned long ul;
typedef unsigned long long ull;
typedef long long ll;
typedef vector<ll> vint;
typedef vector<vector<ll>> vvint;
typedef vector<vector<vector<ll>>> vvvint;
typedef vector<string> vstring;
typedef vector<vector<string>> vvstring;
typedef vector<char> vchar;
typedef vector<vector<char>> vvchar;
typedef vector<long double> vdouble;
typedef vector<vector<long double>> vvdouble;
typedef vector<vector<vector<long double>>> vvvdouble;
typedef pair<ll,ll> pint;
typedef vector<pint> vpint;
typedef vector<bool> vbool;

#define rep(i,n) for(ll i=0;i<n;i++)
#define repf(i,f,n) for(ll i=f;i<n;i++)
#define repr(i,n) for(ll i=n-1;i>=0;i--)
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ALL(obj) (obj).begin(), (obj).end()
// #define LLONG_MAX 9223372036854775806
#define vmax(vec) *max_element(vec.begin(), vec.end())
#define vmin(vec) *min_element(vec.begin(), vec.end())
#define vsort(vec) sort(vec.begin(), vec.end())
#define vsortgr(vec) sort(vec.begin(), vec.end(), greater<ll>())
#define MOD 1000000007
// #define MOD 998244353

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

//
// struct Node{
//   vint children;
//   ll index;
//   ll prop;
// };
// struct edge{ll to; ll cost;};



int dy[]={0, 0, 1, -1};
int dx[]={1, -1, 0, 0,};

template< typename T >
struct edge {
  int src, to;
  T cost;

  edge(int to, T cost) : src(-1), to(to), cost(cost) {}

  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WeightedGraph = vector< Edges< T > >;
using UnWeightedGraph = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;

ll gcd(ll a, ll b)
{
    if (a < b) {
        a ^= b;
        b ^= a;
        a ^= b;
    }

    return b ? gcd(b, a % b) : a;
}

ll lcm(int a, ll b)
{
    return a * b / gcd(a, b);
}




// 繰り返し二乗法
ll power(ll a, ll b){
	ll res=1;
	while(b>0){
		if(b&1) res=res*a%MOD;
		a=a*a%MOD;
		b>>=1;
	}
	return res;
}

const int MAX=510000;

ll fact[MAX], fact_inv[MAX];

void init_fact(ll n){
  fact[0]=1;
  // 階乗の計算
  rep(i, n) fact[i+1]=fact[i]*(i+1)%MOD;
  fact_inv[n]=power(fact[n], MOD-2);
  // 逆元の計算
  for(ll i=n-1;i>=0;i--) fact_inv[i]=fact_inv[i+1]*(i+1)%MOD;
}

ll comb(ll n, ll r){
	return (fact[n]*fact_inv[r])%MOD*fact_inv[n-r]%MOD;
}

ll perm(ll n, ll r){
  return (fact[n]*fact_inv[n-r])%MOD;
}


struct UnionFind {
    vector<ll> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2
    vector<ll> num;
    // vector<ll> dist; // rootまでの距離

    UnionFind(ll N) : par(N), num(N){ //最初は全てが根であるとして初期化
      for(ll i = 0; i < N; i++) par[i] = i;
      for(ll i = 0; i < N; i++) num[i] = 1;
      // for(ll i = 0; i < N; i++) dist[i] = 0;

    }

    ll root(ll 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(ll x, ll y) { // xとyの木を併合
        ll rx = root(x); //xの根をrx
        ll 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
        num[ry] = num[rx] + num[ry];
    }



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

    ll size(ll x){
      return num[root(x)];
    }


};

vint divisor(ll n){ // nの約数
  vint ret;
  for(ll i=1;i*i<=n;i++){
    if(n%i == 0){
      ret.pb(i);
      if(i*i!=n) ret.pb(n/i);
    }
  }
  vsort(ret);
  return ret;
}

ll my_pow(ll x,ll y){
  ll rtn=1;
  rep(i,y) rtn*=x;
  return rtn;
}

ll get_digit_in(ll n, ll i){ // i桁目の数字を得る。
  for(ll j=0;j<i-1;j++){
    n/=10;
  }
  return n%10;
}

ll get_digit(ll n){
  ll rtn=0;
  while(n>0){
    n/=10;
    rtn++;
  }
  return rtn;
}

map<ll,ll> prime_factor(ll n){
  map<ll,ll> rtn;
  for(ll i=2;i*i<=n;i++){
    while(n%i == 0){
      rtn[i]++;
      n /= i;
    }
  }
  if(n!=1) rtn[n] = 1;
  return rtn;
}

// nPnの順列に対して処理を実行する
void foreach_permutation(ll n, function<void(ll *)> f) {
  ll indexes[n];
  for (ll i = 0; i < n; i++) indexes[i] = i;
  do {
    f(indexes);
  } while (std::next_permutation(indexes, indexes + n));
}

void recursive_comb(ll *indexes, ll s, ll rest, function<void(ll *)> f) {
  if (rest == 0) {
    f(indexes);
  } else {
    if (s < 0) return;
    recursive_comb(indexes, s - 1, rest, f);
    indexes[rest - 1] = s;
    recursive_comb(indexes, s - 1, rest - 1, f);
  }
}

// nCkの組み合わせに対して処理を実行する
void foreach_comb(ll n, ll k, function<void(ll *)> f) {
  ll indexes[k];
  recursive_comb(indexes, n - 1, k, f);
}

// nPkの順列に対して処理を実行する
void foreach_permutation(ll n, ll k, function<void(ll *)> f) {
  foreach_comb(n, k, [&](ll *c_indexes) {
    foreach_permutation(k, [&](ll *p_indexes) {
      ll indexes[k];
      for (ll i = 0; i < k; i++) {
        indexes[i] = c_indexes[p_indexes[i]];
      }
      f(indexes);
    });
  });
}

// ll arr[1000500];
// vint primes;

// //エラトステネスの篩
// void Eratosthenes(){
//   ll N = 1000500;
// 	for(ll i = 0; i < N; i++){
// 		arr[i] = 1;
// 	}
// 	for(ll i = 2; i < sqrt(N); i++){
// 		if(arr[i]){
// 			for(ll j = 0; i * (j + 2) < N; j++){
// 				arr[i *(j + 2)] = 0;
// 			}
// 		}
// 	}

// 	for(ll i = 2; i < N; i++){
// 		if(arr[i]){
//       primes.pb(i);
// 			//cout << i << endl;
// 		}
// 	}
// }


void print(vint &V){
  for(auto e:V)cout<<e<<" ";
  cout<<endl;
}





int main() {
  cout<<fixed<<setprecision(12);

  ll a,b;cin>>a>>b;


  cout<<max(a-2*b, 0ll)<<endl;

}