Farmer John has been informed of the location of a fugitive cow and wants to catch her immediately. He starts at a point N (0 ≤ N ≤ 100,000) on a number line and the cow is at a point K (0 ≤ K ≤ 100,000) on the same number line. Farmer John has two modes of transportation: walking and teleporting.

* Walking: FJ can move from any point X to the points X - 1 or X + 1 in a single minute
* Teleporting: FJ can move from any point X to the point 2 × X in a single minute.

If the cow, unaware of its pursuit, does not move at all, how long does it take for Farmer John to retrieve it?

Input

Line 1: Two space-separated integers: N and K

Output

Line 1: The least amount of time, in minutes, it takes for Farmer John to catch the fugitive cow.

Sample Input

5 17

Sample Output

4

Hint

The fastest way for Farmer John to reach the fugitive cow is to move along the following path: 5-10-9-18-17, which takes 4 minutes.
        
题意：

农场主约翰已经得知逃亡母牛的位置，并想立即抓住她。在一条数列上，他在点N（0±n＝100000）处开始，在同一条线上，母牛处于点K（0±k±100000）。农民约翰有两种交通方式：步行和长途运输。


*行走：FJ可以从任意点X移动到点X - 1或X+ 1在一分钟内

*远程传送：FJ可以在一分钟内从任意点X移动到点2×X。


如果奶牛不知道它的追求，它根本不动，农民约翰需要多久才能取回它呢？

思路：

    广搜，如果n>=k，就只能往回行走了，如果n<k，三种走法都可以，利用广搜把符合条件的位置，步数加入队列，并标记为走过了。每次从队首开始查找并更新队首，直到查找到牛的位置，输出步数。

   不要忘记初始化标记的数组！

代码：

#include<stdio.h>
#include<string.h>
int book[100010];
struct data
{
	int x;
	int step;
}q[1000000],t;
int main()
{
	int n,head,tail,k;
	while(scanf("%d%d",&n,&k)!=EOF)
	{
		memset(book,0,sizeof(book));
		if(n>=k)
			printf("%d\n",n-k);
		else
		{
			head=0;
			tail=0;
			book[n]=1;
			q[tail].x=n;
			q[tail].step=0;
			tail++;
			while(head<tail)
			{
				t=q[head++];
				if(t.x+1>0&&t.x+1<100010&&book[t.x+1]==0)
				{
					q[tail].x=t.x+1;
					q[tail].step=t.step+1;
					tail++;
					book[t.x+1]=1;
					if(t.x+1==k)
					{
						printf("%d\n",t.step+1);
						break;
					}
				}
				if(t.x-1>0&&t.x-1<100010&&book[t.x-1]==0)
				{
					q[tail].x=t.x-1;
					q[tail].step=t.step+1;
					tail++;
					book[t.x-1]=1;
					if(t.x-1==k)
					{
						printf("%d\n",t.step+1);
						break;
					}	
				}
				if(t.x*2>0&&t.x*2<100010&&book[t.x*2]==0)
				{
					q[tail].x=t.x*2;
					q[tail].step=t.step+1;
					tail++;
					book[t.x*2]=1;
					if(t.x*2==k)
					{
						printf("%d\n",t.step+1);
						break;
					}
				}	
			}
		}
	}
	return 0;
}
#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
struct note
{
	int x;
	int step;
};
int book[200010];
int main()
{
	struct note A,B,next;
	queue<note>que;
	int a,b,k;
	while(scanf("%d%d",&a,&b)!=EOF)
	{
		memset(book,0,sizeof(book));
		while(!que.empty())
		{
			que.pop();
		}
		A.x=a;
		A.step=0;
		book[A.x]=1;
		que.push(A);
		while(!que.empty())
		{
			B=que.front();
			if(B.x==b)
			{
				printf("%d\n",B.step);
				break;
			}
			que.pop();
			for(k=0;k<3;k++)
			{
				next=B;
				if(k==0)
					next.x=next.x+1;
				if(k==1)
					next.x=next.x-1;
				if(k==2)
					next.x=next.x*2;
				next.step++;
				if(next.x>=0&&next.x<=200000&&book[next.x]==0)
				{
					book[next.x]=1;
					que.push(next);
				}
			}
			
		}
	}
	return 0;
}