Catch That Cow

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.

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
int n;
typedef pair<int,int> tp;
queue <tp> pp;
int sn[200010]= {0};
int bfs(int m,int s)
{
    if(m>=n)
        return m-n;
    pp.push(tp(m,s));
    while(pp.size())
    {
        if(m<0) continue;
        tp p=pp.front();
        pp.pop();
        if(sn[p.first]==0)
        {
            sn[p.first] =1;
            if(p.first==n) return p.second;
            if(p.first+1<=1+n)
                pp.push(tp(p.first+1,p.second+1));
            if(p.first-1<=1+n)
                pp.push(tp(p.first-1,p.second+1));
            if(p.first*2<=1+n&&p.first!=0)
                pp.push(tp(p.first*2,p.second+1));
        }
    }
    return 0;
}
int main()
{
    int m;
    scanf("%d%d",&m,&n);
    printf("%d",bfs(m,0));
    return 0;
}