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++ (GCC 9.2.1) (judge verdict: Accepted)

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
#include<deque>
using namespace std;
#define int long long
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define UF UnionFind 
//printf("%.12Lf\n",);
int keta(int x) {
	rep(0, i, 30) {
		if (x < 10) {
			return i + 1;
		}
		x = x / 10;
	}
}
int gcd(int x, int y) {
	if (x == 0 || y == 0)return x + y;
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return bb;
		}
		bb = bb % aa;
		if (bb == 0) {
			return aa;
		}
	}
}
int lcm(int x, int y) {
	int aa = x, bb = y;
	rep(0, i, 1000) {
		aa = aa % bb;
		if (aa == 0) {
			return x / bb * y;
		}
		bb = bb % aa;
		if (bb == 0) {
			return x / aa * y;
		}
	}
}
bool p(int x) {
	if (x == 1)return false;
	rep(2, i, sqrt(x) + 1) {
		if (x % i == 0 && x != i) {
			return false;
		}
	}
	return true;
}
int max(int a, int b) {
	if (a >= b)return a;
	else return b;
}
string maxst(string s, string t) {
	int n = s.size();
	int m = t.size();
	if (n > m)return s;
	else if (n < m)return t;
	else {
		rep(0, i, n) {
			if (s[i] > t[i])return s;
			if (s[i] < t[i])return t;
		}
		return s;
	}
}
int min(int a, int b) {
	if (a >= b)return b;
	else return a;
}
int n2[61];
int nis[61];
int nia[61];
int mody[61];
int nn;
int com(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int bunsi = 1, bunbo = 1;
	rep(0, i, y)bunsi = (bunsi * (n - i)) % mod;
	rep(0, i, y)bunbo = (bunbo * (i + 1)) % mod;
	mody[0] = bunbo;
	rep(1, i, 41) {
		bunbo = (bunbo * bunbo) % mod;
		mody[i] = bunbo;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			bunsi = (bunsi * mody[i]) % mod;
		}
	}
	return bunsi;
}
int gyakugen(int n, int y) {
	int ni = 1;
	for (int i = 0;i < 41;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	mody[0] = y;
	rep(1, i, 41) {
		y = (y * y) % mod;
		mody[i] = y;
	}
	rep(0, i, 41)nis[i] = 0;
	nn = mod - 2;
	for (int i = 40;i >= 0;i -= 1) {
		if (nn > n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	nis[0]++;
	rep(0, i, 41) {
		if (nis[i] == 1) {
			n = (n * mody[i]) % mod;
		}
	}
	return n;
}
int yakuwa(int n) {
	int sum = 0;
	rep(1, i, sqrt(n + 1)) {
		if (n % i == 0)sum += i + n / i;
		if (i * i == n)sum -= i;
	}
	return sum;
}
int poow(int y, int n) {
	if (n == 0)return 1;
	n -= 1;
	int ni = 1;
	for (int i = 0;i < 61;i++) {
		n2[i] = ni;
		ni *= 2;
	}
	int yy = y;
	mody[0] = yy;
	rep(1, i, 61) {
		yy = (yy * yy) % mod;
		mody[i] = yy;
	}
	rep(0, i, 61)nis[i] = 0;
	nn = n;
	for (int i = 60;i >= 0;i -= 1) {
		if (nn >= n2[i]) {
			nis[i]++;
			nn -= n2[i];
		}
	}
	rep(0, i, 61) {
		if (nis[i] == 1) {
			y = (y * mody[i]) % mod;
		}
	}
	return y;
}
int minpow(int x, int y) {
	int sum = 1;
	rep(0, i, y)sum *= x;
	return sum;
}
int ketawa(int x, int sinsuu) {
	int sum = 0;
	rep(0, i, 100)sum += (x % poow(sinsuu, i + 1)) / (poow(sinsuu, i));
	return sum;
}
int sankaku(int a) {
	return a * (a + 1) / 2;
}
int sames(int a[1111111], int n) {
	int ans = 0;
	rep(0, i, n) {
		if (a[i] == a[i + 1]) {
			int j = i;
			while (a[j + 1] == a[i] && j <= n - 2)j++;
			ans += sankaku(j - i);
			i = j;
		}
	}
	return ans;
}
using Graph = vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph& G, int v, int p, int d) {
	depth[v] = d;
	oya[v] = p;
	for (auto nv : G[v]) {
		if (nv == p) continue; // nv が親 p だったらダメ
		dfs(G, nv, v, d + 1); // d を 1 増やして子ノードへ
	}
}
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
	memo[h][w] = 'x';

	// 八方向を探索
	for (int dh = -1; dh <= 1; ++dh) {
		for (int dw = -1; dw <= 1; ++dw) {
			if(abs(0-dh)+abs(0-dw)==2)continue;
			int nh = h + dh, nw = w + dw;

			// 場外アウトしたり、0 だったりはスルー
			if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
			if (memo[nh][nw] == 'x') continue;

			// 再帰的に探索
			dfs(nh, nw);
		}
	}
}*/
int XOR(int a, int b) {
	if (a == 0 || b == 0) {
		return a + b;
	}
	int ni = 1;
	rep(0, i, 41) {
		n2[i] = ni;
		ni *= 2;
	}
	rep(0, i, 41)nis[i] = 0;
	for (int i = 40;i >= 0;i -= 1) {
		if (a >= n2[i]) {
			nis[i]++;
			a -= n2[i];
		}
		if (b >= n2[i]) {
			nis[i]++;
			b -= n2[i];
		}
	}
	int sum = 0;
	rep(0, i, 41)sum += (nis[i] % 2 * n2[i]);
	return sum;
}
int string_int(string s){
	int sum=0;
	int n=s.size();
	int t=1;
	rep(0,i,n){
		sum+=t*(s[n-i-1]-'0');
		t*=10;
	}	
 	 return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=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 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;
	}
};
signed main(){
	ic(a) ic(b)
	c(max(0,a-b*2))
}