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;}