Time limit                   2000 msMemory          limit         65536 kB

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

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.

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int n,k;
int step[maxn];
bool vis[maxn];
queue<int>Q;
int bfs()
{
    int head,next;
    Q.push(n);
    step[n]=0;
    vis[n]=true;
    while(!Q.empty())
    {
        head=Q.front();
        Q.pop();
        for(int i=0;i<3;i++)
        {
            if(i==0) next=head-1;
            else if(i==1) next=head+1;
            else next=head*2;
            if(next<0||next>=maxn) continue;
            if(!vis[next])
            {
                Q.push(next);
                step[next]=step[head]+1;
                vis[next]=true;
            }
            if(next==k) return step[next];
        }
    }

}
int main()
{
    while(scanf("%d%d",&n,&k)==2)
    {
        if(n>=k) printf("%d\n",n-k);
        else
        {
            while(!Q.empty()) Q.pop();
            memset(vis,false,sizeof(vis));
            memset(step,0,sizeof(step));
            printf("%d\n",bfs());
        }
    }
    return 0;
}