Problem Description

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.

讲解：一农夫追牛，牛很笨，他就在原地等着农夫来抓他，并且这是一条直线，农夫很容易就能找到它的，然而农夫却只有三种选择，退一步，走一步，或者走到当前位置的2倍；于是乎我们可以用搜索来解决；

代码如下：

1 #include
       
         2 #include
        
          3 #include
         
           4 #include
          
            5 #include
           
             6 using namespace std; 7 int n,m; 8 int map[200010]; 9 struct T10 {11     int x,step;12 };13 int dfs(T now)14 {queue< T >q;15     T end;16     q.push(now);17     while(!q.empty())18     {19         end=q.front();20           q.pop();21         map[end.x]=1;22     if(end.x==m)23      {
     return end.step;}24      //如果不是走一步一判断很容易超出内存的25      now.x=end.x+1;//前进一步，存入队列；26          if(end.x>=0 && end.x<=100000 && map[now.x]==0)27          {28              now.step=end.step+1;29              map[now.x]==1;30              q.push(now);31          }32      now.x=end.x-1;//后退一步33      if(end.x>=0 && end.x<=100000 && map[now.x]==0)34      {35          now.step=end.step+1;36          map[now.x]==1;37          q.push(now);38      }39      now.x=end.x*2;//前进2倍的位置40      if(end.x>=0 && end.x<=100000 && map[now.x]==0)41      {42          now.step=end.step+1;43          map[now.x]==1;44          q.push(now);45      }46     }47     return 0;48 }49 int main()50 {51     T now;52     while(cin>>n>>m)53     {54         if(n>=m)//如果n大于m则只能后退了；55        {56            cout<
            
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