Polycarp plays “Game 23”. Initially he has a number n and his goal is to transform it to m. In one move, he can multiply n by 2 or multiply n by 3. He can perform any number of moves.
Print the number of moves needed to transform n to m. Print -1 if it is impossible to do so.
It is easy to prove that any way to transform n to m contains the same number of moves (i.e. number of moves doesn’t depend on the way of transformation).
Input
The only line of the input contains two integers n and m (1≤n≤m≤5⋅108).
Output
Print the number of moves to transform n to m, or -1 if there is no solution.
Examples
Input
120 51840
Output
7
Input
42 42
Output
0
Input
48 72
Output
-1
Note
In the first example, the possible sequence of moves is: 120→240→720→1440→4320→12960→25920→51840. The are 7 steps in total.
In the second example, no moves are needed. Thus, the answer is 0.
In the third example, it is impossible to transform 48 to 72.
题意：该题的思路是要用m乘2或乘3一直到n;算出一共乘几步；
题解：因为n是用2和3从m乘起来的，所以先用n/m,如果不能整除，说明就是“-1”，如果能，说明商/2或/3就是总共的步数；
#include<cstdio>
using namespace std;
int main()
{
long long a,b; a代表题中n，b代表题中m。
while(~scanf("%lld%lld",&a,&b)){
long long x = 0; //x记录执行的次数
if(b%a != 0){
printf("-1\n");
}
else{
b /= a;
while(b != 1){
/*if(b%6 == 0){ // 2x3 = 6，可以直接留下光剩下2或者3的情况，也可以不加，直接忽略
b /= 6;
x += 2;
}*/
if(b%3 == 0){
b /= 3;
x++;
}
else if(b%2 == 0){
b /= 2;
x++;
}
else{ //例如 1 5这种数据在前面b/a的时候无法筛掉，所以在这直接break然后跳出
x = -1;
break;
}
}
printf("%lld\n",x);
}
}
return 0;
}