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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：

 农夫在n点，牛在k点，找到从n到k的最短时间。

 bfs代码：

#include
   
    #include
    
     #include
     
      #include
      
       #include
       
        using namespace std;int n,k;struct node{	int x;    //位置	int step; //时间};int vis[200000];  //标记走过的点queue
        
          q;void bfs(){ while(!q.empty())q.pop(); //队列置空 memset(vis,0,sizeof(vis)); vis[n]=1; struct node nod={n,0}; q.push(nod); while(!q.empty()) { struct node newn=q.front(); q.pop(); int tx; if(newn.x==k) //如果现在的位置=牛所在的位置就停止循环 { printf("%d\n",newn.step); break; } for(int i=0;i<3;i++) { switch(i){ case 0: tx=newn.x+1; break; case 1: tx=newn.x-1; break; case 2: tx=newn.x*2; break; } if(tx>=0&&tx<=200000&&vis[tx]==0) //所在的位置符合条件就入队 { struct node d={tx,newn.step+1}; q.push(d); vis[tx]=1; } } } return;}int main(){ while(scanf("%d%d",&n,&k)!=EOF) { bfs(); } return 0;}
        
       
      
     
    
   

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