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.

結論：k進位制下的n的階乘的位數=logk(n)=log10(n)/(log10(k))，所以可以先求出10進位制下f[n]的對數打表。另外f[n]應先保持浮點型，最後向上取整

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include<cmath>
#define MAXN 1000010
using namespace std;
double a[MAXN];
void Init()
{
    a[0]=log10(1);
    for(int i=1;i<=1000000;++i)
        a[i]=a[i-1]+log10(i);
}
int main()
{
    Init();
    int ans;
    int t,cnt=1,i,j,m,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        if(n==0)
            ans=1;
        else
        {
            ans=ceil(a[n]/log10(m));

        }
        printf("Case %d: %d\n",cnt++,ans);
    }
    return 0;
}