描述：

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.

題意：

       農民和奶牛各在一個地方，農民可以通過+1，-1，*2來移動，每走一步用時一分鐘，問移動到奶牛所在的位置需要多久。

題解：

         廣搜，分別進行+1，-1，*2操作，存入佇列。

程式碼：

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>

using namespace std;
queue<int>q;
int a,b;
bool vis[100002];
int step[100002];

int bfs()
{
	int u,v;
	q.push(a);
	step[a]=0;
	vis[a]=true;
	while(!q.empty())
	{
		u=q.front();
		q.pop();
		for(int i=0;i<3;i++)
		{
			if(i==0) v=u+1;
			else if(i==1) v=u-1;
			else v=u*2;
			if(v>=100001||v<0) continue;
			if(!vis[v])
			{
				step[v]=step[u]+1;
				vis[v]=true;
				q.push(v);
			}
			if(v==b) return step[v];
		}
	}
	return -1;
}

int main()
{
	while(cin>>a>>b)
	{
		memset(step,0,sizeof(step));
		memset(vis,false,sizeof(vis));
		while(!q.empty()) q.pop();
		if(a>=b) printf("%d\n",a-b);
		else
		{
			int ans=bfs();
			cout<<ans<<endl;
		}
	}
	return 0;
}