C

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<iostream>#include<queue>#include<cstring>using namespace std;#define Max 100000int visited[Max];struct Way{int x;int time;Way(int x1,int time1){x=x1;time=time1;}};int main(){ios::sync_with_stdio(false);int N,K;memset(visited,0,sizeof(visited));queue <Way> cow;cin>>N>>K;cow.push(Way(N,0));visited[N]=1;while(!cow.empty()){Way s=cow.front();if(s.x==K){cout<<s.time<<endl;break;}else{if(s.x-1>=0&&!visited[s.x-1])//左移一位 {cow.push(Way(s.x-1,s.time+1));visited[s.x-1]=1;}if(s.x+1<=Max&&!visited[s.x+1])//右移一位 {cow.push(Way(s.x+1,s.time+1));visited[s.x+1]=1;}if(s.x*2<=Max&&!visited[s.x*2])//直接到达2*x点 {cow.push(Way(s.x*2,s.time+1));visited[s.x*2];}}cow.pop();}return 0;}