Factorial of an integer is defined by the following function

f(0) = 1

f(n) = f(n - 1) * n, if(n > 0)

So, factorial of 5 is 120. But in different bases, the factorial may be different. For example, factorial of 5 in base 8 is 170.

In this problem, you have to find the number of digit(s) of the factorial of an integer in a certain base.

Input

Input starts with an integer T (≤ 50000), denoting the number of test cases.

Each case begins with two integers n (0 ≤ n ≤ 106) and base (2 ≤ base ≤ 1000). Both of these integers will be given in decimal.

Output

For each case of input you have to print the case number and the digit(s) of factorial n in the given base.

Sample Input

5

5 10

8 10

22 3

1000000 2

0 100

Sample Output

Case 1: 3

Case 2: 5

Case 3: 45

Case 4: 18488885

Case 5: 1

#include<cmath>
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<cstring>
#define maxn 1000010
#define ll long long
using namespace std;
int t;
double a[maxn];
ll n,k;
double ans;
void init()
{
    a[0]=log(1);
    for(int i=1;i<maxn;i++)
       a[i]=a[i-1]+log(i);
}
int main()
{
    scanf("%d",&t);
    init();
    int w=0;
    while(t--)
    {w++;
ll n,k;
        scanf("%lld%lld",&n,&k);
     double ans=a[n]/log(k)+1;
int sum=(int)ans;
        printf("Case %d: %d\n",w,sum);
    }
    return 0;

}