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 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>
using namespace std;

struct node{
    int loc, step;//loc 当前位置， step 次数
}pos,q;

queue<node> que;
int vis[200005];

void Push(int loc, int step){
    q.loc = loc;
    q.step = step + 1;
    vis[loc] = 1;
    que.push(q);
}

int bfs(int n, int k){
    pos.loc = n, pos.step = 0;
    vis[n] = 1;
    que.push(pos);
    while(!que.empty()){
        pos = que.front();
        que.pop();
        if(pos.loc == k)
            return pos.step;
        int loc_mov = pos.loc - 1;
        if(loc_mov >= 0 && loc_mov < 200005 && vis[loc_mov] == 0){
            Push(loc_mov, pos.step);
        }
        loc_mov = pos.loc + 1;
        if(loc_mov >= 0 && loc_mov < 200005 && vis[loc_mov] == 0){
            Push(loc_mov, pos.step);
        }
        loc_mov = 2 * pos.loc;
        if(loc_mov >= 0 && loc_mov < 200005 && vis[loc_mov] == 0){
            Push(loc_mov, pos.step);
        }
    }
    return -1;
}

int main(){
    int n, k;
    cin >> n >> k;
    cout << bfs(n, k) << endl;
}