CodeForces

Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows numbern.

Bob wants to get number m on the display. What minimum number of clicks he has to make in order to achieve this result?

Input

The first and the only line of the input contains two distinct integers n and m (1 ≤ n, m ≤ 10⁴), separated by a space .

Output

Print a single number — the minimum number of times one needs to push the button required to get the numberm out of number n.

Example

Input

4 6

Output

2

Input

10 1

Output

9

Note

In the first example you need to push the blue button once, and then push the red button once.

In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.

题意：有两个键，一个使数字*2，一个使数字-1，给你两个数，让你把第一个数变成第二个数，问你最少的操作使多少次。

思路：广搜枚举，标记，但是要注意的是，数据范围，当你枚举出来的数大于数据范围，就不要继续下去了

include<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;int n,m,f;struct node{int x;int stemp;}u,e;int book[1000000];int text(int x){if(x<=0 || x>131072)//个人认为只要大于2*10^4就不需要继续枚举了，比赛的时候按照2的幂得出的这个数    return 0;return 1;}void BFS(){    queue<node>q;    while(!q.empty()) q.pop();    u.x=n;    u.stemp=0;    q.push(u);    book[u.x]=1;    while(!q.empty()){        u=q.front();        q.pop();     //   printf(" C u.x=%d u.stemp=%d\n",u.x,u.stemp);        if(u.x==m){           printf("%d\n",u.stemp);           break;        }        e.x=u.x-1;        e.stemp=u.stemp+1;        if(text(e.x)==1 && book[e.x]==0){            q.push(e);            book[e.x]=1;         //   printf(" R e.x=%d e.stemp=%d\n",e.x,e.stemp);        }        e.x=u.x*2;        if(text(e.x)==1 && book[e.x]==0){            q.push(e);             book[e.x]=1;          //   printf(" R e.x=%d e.stemp=%d\n",e.x,e.stemp);        }    }}int main(){    scanf("%d%d",&n,&m);    memset(book,0,sizeof(book));    f=0;    BFS();}