LightOJ

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 andn 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

Case 2: 6

Case 3: 12

代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    int r=1;
    while(t--)
    {
        long long n,k;
        scanf("%lld%lld",&n,&k);
        int ans=1;
        long long s=k;
        while(s%n!=0)
        {
            s=(s*10+k)%n;
            ans++;
        }
     printf("Case %d: %d\n",r++,ans);
    }
}



#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    int r=1;
    while(t--)
    {
        long long n,k;
        scanf("%lld%lld",&n,&k);
        int ans=1;
        long long s=k;
        while(s%n!=0)
        {
            s=(s*10+k)%n;
            ans++;
        }
     printf("Case %d: %d\n",r++,ans);
    }
}