D

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status

Description

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.

Sample Input

3

3 1

7 3

9901 1

Sample Output

#include<cstdio>
#include<cmath>
#include<cstring>
int main()
{
    int t,i,j,ans,num,n,digit;
    scanf("%d",&t);
    for(i = 1;i <= t;i++)
    {
        scanf("%d %d",&n,&digit);
        ans = digit ;
        j = 1;
        num =1;
        while(1)
        {
            if(ans%n!=0)
            {
                ans=ans%n*10+digit;
                j++;
                num++;    
            }
            if(ans%n==0)
            {
                break;
            }
        }
        printf("Case %d: %d\n",i,num);
    }
    return 0;
}

 

Case 1: 3

Case 2: 6

Case 3: 12