B. Fox Dividing Cheese----GCD/因子问题

B. Fox Dividing Cheese

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Two little greedy bears have found two pieces of cheese in the forest of weight a and b grams, correspondingly. The bears are so greedy that they are ready to fight for the larger piece. That's where the fox comes in and starts the dialog: "Little bears, wait a little, I want to make your pieces equal" "Come off it fox, how are you going to do that?", the curious bears asked. "It's easy", said the fox. "If the mass of a certain piece is divisible by two, then I can eat exactly a half of the piece. If the mass of a certain piece is divisible by three, then I can eat exactly two-thirds, and if the mass is divisible by five, then I can eat four-fifths. I'll eat a little here and there and make the pieces equal".

The little bears realize that the fox's proposal contains a catch. But at the same time they realize that they can not make the two pieces equal themselves. So they agreed to her proposal, but on one condition: the fox should make the pieces equal as quickly as possible. Find the minimum number of operations the fox needs to make pieces equal.

Input

The first line contains two space-separated integers a and b (1 ≤ a, b ≤ 109).

Output

If the fox is lying to the little bears and it is impossible to make the pieces equal, print -1. Otherwise, print the required minimum number of operations. If the pieces of the cheese are initially equal, the required number is 0.

Examples

input

15 20

output

3

input

14 8

output

-1

input

6 6

output

0

题目链接：http://codeforces.com/contest/371/problem/B

题目的意思是给你两个数a，b，这两个数可以除以1/2,1/3,1/5，问最少变换多少次才能使两个数相等。

我猜的和gcd有关，没想到还真的有关系。

我们来看第一组示例，15=3*5,20=2*2*5,去掉公因数5，还剩下3个数，输出3即可，不能被2,3,5整除的输出-1.

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <cmath>#include <algorithm>using namespace std;int c[4],d[4];int main(){    int a,b;    scanf("%d%d",&a,&b);    if(a==b){        printf("0\n");        return 0;    }    int gcd=__gcd(a,b);    a/=gcd;    b/=gcd;    while(a%2==0){        a/=2;        c[0]++;    }    while(a%3==0){        a/=3;        c[1]++;    }    while(a%5==0){        a/=5;        c[2]++;    }    while(b%2==0){        b/=2;        d[0]++;    }    while(b%3==0){        b/=3;        d[1]++;    }    while(b%5==0){        b/=5;        d[2]++;    }    if(a!=b){        printf("-1\n");        return 0;    }    //for(int i=0;i<3;i++){     //   printf("%d %d\n",c[i],d[i]);    //}    int ans=0;    for(int i=0;i<3;i++){        ans+=c[i];        ans+=d[i];    }    printf("%d\n",ans);}

所谓神题必有神解，回滚做法http://blog.csdn.net/synapse7/article/details/21085153