Find a minimal interger K which is merely comprised of N and can be divided by M.

For example,11 is the minimal number that and be divided by 11, and it is comprised of two '1's, and 111111 can be divided by 13 which is comprised of six '1's.

Input

On each line of input , there will be two positive integer, N and M. N is a digit number, M is no more than 10000.

Output

On each single line, output the number of N, if no such K, output zero.

Sample Input

1 5
1 11
1 13

Sample Output

0
2
6

据说暴力就可以过了，不过我还是用了数论中的东西

不过个人感觉这个题要么数据太强大，要么就是数据有问题，我代码中有一个无关紧要的东西，居然改掉了就错了。。我也不知道为啥

至于做法，可以参考POJ上的一个题目，当然那个题我也写了相应的解题报告

http://poj.org/problem?id=3696

我的代码：

#include<stdio.h>
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<math.h>

using namespace std;

typedef long long ll;

ll gcd(ll a,ll b)
{
	if(b==0)
		return a;
	else
		return gcd(b,a%b);
}

ll eular(ll n)
{
	ll res=1,i;
	for(i=2;i*i<=n;i++)
	{
		if(n%i==0)
		{
			res=res*(i-1);
			n=n/i;
			while(n%i==0)
			{
				res=res*i;
				n=n/i;
			}
		}
		if(n==1)
			break;
	}
	if(n>1)
		res=res*(n-1);
	return res;
}

ll exmod(ll p,ll n,ll m)
{
	ll sq=1;
	while(n>0)
	{
		if(n%2==1)
			sq=(sq%m)*(p%m)%m;
		p=(p%m)*(p%m)%m;
		n=n/2;
	}
	return sq%m;
}

int main()
{
	ll n,m,g,M,phi,i;
	ll a[1000],num;
	while(scanf("%lld%lld",&n,&m)!=EOF)
	{
		g=gcd(m,n);//这里理论上取9m和n的最大公约数也是可行的。恩，但是改掉就WA了，我也不知道为啥
		M=9*m/g,num=0;
		if(gcd(M,10)!=1)
		{
			printf("0\n");
			continue;
		}
		phi=eular(M);
		for(i=1;i*i<=phi;i++)//其实这里还可以更快，直接质因数分解之后来搞，恩
		{
			if(phi%i==0)
			{
				a[num++]=i;
				a[num++]=phi/i;
			}
			if(i*i==phi)
				a[num++]=i;
		}
		sort(a,a+num);
		for(i=0;i<num;i++)
			if(exmod(10,a[i],M)==1)
			{
				printf("%lld\n",a[i]);
				break;
			}
	}
	return 0;
}