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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 - 1 or + 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

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1
5 17

Sample Output

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1
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.

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#include<iostream> #include<cmath> #include<string.h> #include<algorithm> #include<cstdio> #include<queue> #include<set> #include<map> #include<vector> #include<string> using namespace std; typedef long long ll; int vis[100001]={0}; int n,m; void bfs(int start) { queue<int>q; q.push(start); int head; while(!q.empty()) { head = q.front(); q.pop(); if(head == m) { break; } if(head>0&&!vis[head-1]) { int New = head-1; vis[New] = vis[head]+1; q.push(New); } if(head<m&&!vis[head+1]) { int New = head+1; vis[New] = vis[head]+1; q.push(New); } if(2*head<100001&&!vis[2*head]) { int New = 2*head; vis[New] = vis[head]+1; q.push(New); } } } int main(){ while(cin>>n>>m) { memset(vis,0,sizeof(vis)); bfs(n); cout<<vis[m]<<endl; } }

 

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