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.
 
 别忘记剪枝啊！要不然超时和超内存多难受TAT
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 #include<vector>
 5 #include<cstring>
 6 #include<string>
 7 #include<algorithm>
 8 #include<map>
 9 #include<cmath>
10 #include<math.h>
11 using namespace std;
12 
13 int n,k,res;
14 bool vis[200005];
15 
16 struct node
17 {
18     int p,t;//p是当前位置,t是所用时间
19 }a;
20 
21 void bfs()
22 {
23     node tmp,next;
24     queue<node>Q;
25     Q.push(a);
26     while(!Q.empty())
27     {
28         tmp=Q.front();
29         Q.pop();
30         if(tmp.p==k)
31         {
32             if(tmp.t<res)
33                 res=tmp.t;
34         }
35         else if(tmp.t>res)
36             continue;
37         else 
38         {
39             next.t=tmp.t+1;
40             //往前
41             if(tmp.p<=k&&!vis[tmp.p+1])
42             {
43                 next.p=tmp.p+1;
44                 Q.push(next);
45                 vis[next.p]=true;
46             }
47             //往后
48             if(tmp.p>=1&&!vis[tmp.p-1])
49             {
50                 next.p=tmp.p-1;
51                 Q.push(next);
52                 vis[next.p]=true;
53             }
54             //乘2
55             if(tmp.p<=k&&!vis[tmp.p*2])
56             {
57                 next.p=tmp.p*2;
58                 Q.push(next);
59                 vis[next.p]=true;
60             }
61             
62         }
63         
64     }
65 }
66 
67 int main()
68 {
69     while(~scanf("%d%d",&n,&k))
70     {
71         res=abs(k-n);
72         a.p=n;
73         a.t=0;
74         memset(vis,false,sizeof(vis));
75         vis[n]=true;
76         bfs();
77         printf("%d\n",res);
78     }
79     return 0;
80 }