If an integer is not divisible by 2 or 5, some multiple of that number in decimal notation is a sequence of only a digit. Now you are given the number and the only allowable digit, you should report the number of digits of such multiple.

For example you have to find a multiple of 3 which contains only 1's. Then the result is 3 because is 111 (3-digit) divisible by 3. Similarly if you are finding some multiple of 7 which contains only 3's then, the result is 6, because 333333 is divisible by 7.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case will contain two integers n (0 < n ≤ 10⁶ and n will not be divisible by 2 or 5) and the allowable digit (1 ≤ digit ≤ 9).

Output

For each case, print the case number and the number of digits of such multiple. If several solutions are there; report the minimum one.

题意：给出 n和m，让判断多少位的m可以整除n（每一位的数值都为m，如果当前的位数不能整除n则再加上一位）如：3   1意思是多少个1可以整除3显然111可以整除3，所以就是三位，输出3

小技巧：令ans=m 让ans对n取余为ans（覆盖原来的值），每次让ans=ans*10+m（因为不能整除所以增加位数）直到可以整除   原理大概是除法的同余定理吧

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<algorithm>

#include<math.h>

#define LL long long

#define DD double

#define MAX 10000

using namespace std;

int main()

{

    int t;

    int n,m,j,i;

    LL ans;

    scanf("%d",&t);

    int k=1;

    while(t--)

    {

        scanf("%d%d",&n,&m);

        printf("Case %d: ",k++);

        ans=m;

        int sum=1;

        while(ans%n)

        {

            ans=ans%n;

            ans=ans*10+m;

            sum++;

        }

        printf("%d\n",sum);

    }

    return 0;

}