Two Buttons

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.

Examples

Input
4 6
Output
2

Input
10 1
Output
9

经验之谈：（x-1）* 2 * 2 * 2=x * 2 * 2 * 2 - 2 * 2 * 2；
（（x-1）* 2 * 2 - 1 ）* 2=（x-1）* 2 * 2 * 2 - 2；
所以我们把n乘二放大到第一个大于m的数
然后确定应该在哪一次乘二之前减一（不一定一次）直到差值等于0（类似于进制转化）

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int n, m, i, j=0, x=0,k;//x乘2的次数，j减1的次数
	cin >> n >> m;
	while (n < m) {
		n *=2;
		x++;
	}
	k = n - m;
	int x1 = x;
	while (k != 0)
	{
		if (pow(2, x1) <= k) j += k / (int)pow(2, x1), k -= (k / (int)pow(2, x1)) * (int)pow(2, x1);
		else  x1--;
	}
	cout << x + j;
}

疑虑：这是最少的次数吗？
首先x的值，假如小于这个值n一定小于m（初始n<m的前提下）
其次是j的值，从pow（2，x）开始试，先减大值再依次递减