Trailing Zeroes (III)

   You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer Q (1 ≤ Q ≤ 10⁸) in a line.

Output

For each case, print the case number and N. If no solution is found then print 'impossible'.

Sample Input

3

1

2

5

Sample Output

Case 1: 5

Case 2: 10

Case 3: impossible

题意：给出数字，代表某个数的阶乘末尾连续0的个数，求出这个数是多少

代码：

#include<stdio.h>
int num(int n) //求n的阶乘末尾连续0的个数 
{              //百度说是定理 记住吧...
    
 5     int ans=0;
 6     while(n)
 7     {ans+=n/5;
 8      n=n/5;
 9         
10     }
    return ans;    
      
}
int main()
{
    int mid,i=1;
    int t ,q;
    long long l,r;
    scanf("%d",&t);
    while(t--)
    {scanf("%d",&q);
     l=0;
     r=100000000000000;  //r要大于1e8
     long long m=0;
     while(l<=r)
       {mid=(l+r)/2;
        if(num(mid)==q)
          {r=mid-1;
           m=mid;    
          }
        else 
          {if(num(mid)>q)
               r=mid-1;
           else l=mid+1;      
          }  
           
       }
     
      if(m==0) printf("Case %d: impossible
",i); 
      else printf("Case %d: %lld
",i,m);
      i++;    
    }
    return 0;
}