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：

1 #include
     
       2 int num(int n) //求n的阶乘末尾连续0的个数  3 {              //百度说是定理 记住吧... 4      5     int ans=0; 6     while(n) 7     {ans+=n/5; 8      n=n/5; 9         10     }11     return ans;    12       13 }14 int main()15 {16     int mid,i=1;17     int t ,q;18     long long l,r;19     scanf("%d",&t);20     while(t--)21     {scanf("%d",&q);22      l=0;23      r=100000000000000;  //r要大于1e824      long long m=0;25      while(l<=r)26        {mid=(l+r)/2;27         if(num(mid)==q)28           {r=mid-1;29            m=mid;    30           }31         else 32           {
   if(num(mid)>q)33                r=mid-1;34            else l=mid+1;      35           }  36            37        }38      39       if(m==0) printf("Case %d: impossible\n",i); 40       else printf("Case %d: %lld\n",i,m);41       i++;    42     }43     return 0;44 }