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=;
     int t ,q;
     long long l,r;
     scanf("%d",&t);
     while(t--)
     {scanf("%d",&q);
      l=;
      r=;  //r要大于1e8
      long long m=;
      while(l<=r)
        {mid=(l+r)/;
         if(num(mid)==q)
           {r=mid-;
            m=mid;
           }
         else
           {if(num(mid)>q)
                r=mid-;
            else l=mid+;
           }  

        }

       if(m==) printf("Case %d: impossible\n",i);
       else printf("Case %d: %lld\n",i,m);
       i++;
     }
     return ;
 }