light oj1078

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

Case 1: 3   111能被3正除

Case 2: 6    333333能被7整除

Case 3: 12      111111111111能被9901整除

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#include<cstdio>   
int main()  
{  
    int n,m,t,k=1;  
    scanf("%d",&t);  
    while(t--){  
        scanf("%d%d",&n,&m);  
        long long ans=1,num=m;  
        while(num%n!=0){  
            num=num*10+m;  
            num%=n;ans++;  
        }  
        printf("Case %d: %lld\n",k++,ans);  
    }  
    return 0;  
}