Catch That Cow

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

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.

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#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;
struct node{
	int x;
	int time;
};
int sx,ex;
int vis[100001];
int b[100001];//没用但是必不可少，缺了就会报runtime error
int bfs()
{
	queue<node>Q;
	node now,next;
	now.x=sx;
	now.time=0;
	Q.push(now);
	while(!Q.empty()){
		now=Q.front();
		if(now.x==ex)
		return now.time;
		if(now.x>0&&!b[now.x-1]){
			next.x=now.x-1;
			next.time=now.time+1;
			b[next.x]=1;
			Q.push(next);	
		}
		if(now.x<ex&&!b[now.x+1]){
			next.x=now.x+1;
			next.time=now.time+1;
			b[next.x]=1;
			Q.push(next);	
		}
		if(now.x<ex&&!b[now.x*2]){
			next.x=now.x*2;
			next.time=now.time+1;
			b[next.x]=1;
			Q.push(next);	
		}
		Q.pop();		
	}
}
int main(){
	
	cin>>sx>>ex;
	b[sx]=1;
	if(sx>=ex)
		cout<<sx-ex;
	else
	cout<<bfs();
	return 0;
}