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Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number n.

Bob wants to get number m on the display. What minimum number of clicks he has to make in order to achieve this result?

Input

The first and the only line of the input contains two distinct integers n and m (1 ≤ n, m ≤ 10⁴), separated by a space .

Output

Print a single number — the minimum number of times one needs to push the button required to get the number m out of number n.

Example

Input

4 6

Output

2

Input

10 1

Output

9

Note

In the first example you need to push the blue button once, and then push the red button once.

In the second example, doubling the number is unnecessary, so we need to push the blue button nine times.

题目分析 ： 有两种操作可以执行，一是将所给的树扩大2倍，2是将所给的数字减 1 ， 之前单纯的写了一个过搜，run time 了....... 后来一位 大佬告诉了我，这个题得剪枝，不然你所开的数组就是会越界。

代码示例 ：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <time.h>
using namespace std;
const int eps = 1e5+5;
const int maxn = 0x3f3f3f3f;
#define ll long long
int n, m;
ll bu[eps];
queue<int>que;

void bfs() { 
    que.push(n);
    
    bu[n] = 0;
    while(!que.empty()) {
        int p = que.front();
        que.pop();
        if (p == m) break;
        int x1 = p*2;
        int x2 = p-1;
        if (bu[x1] == -1 && x1 < 20000) {
            que.push(x1);
            bu[x1] = bu[p] + 1;
        }
        if (bu[x2] == -1 && x2 < 20000) {
            que.push(x2);
            bu[x2] = bu[p] + 1;
        }
    }
}

int main() {
    
    scanf("%d%d", &n, &m);
    memset(bu, -1, sizeof(bu));
    //memset(vis, false, sizeof(vis));
    bfs();
    printf("%lld
", bu[m]);
    
    return 0;
}