C

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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

提示：

令f(x)表示正整数x末尾所含有的“0”的个数，则有：
      当0 < n < 5时，f(n!) = 0;
      当n >= 5时，f(n!) = k + f(k!), 其中 k = n / 5（取整）。 

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
long long sum(long long mid)
{long long sum1=0;
while(mid)
{
sum1=sum1+mid/5;
mid=mid/5;
}
return sum1;

}
int main()
{int t,k=0;
scanf("%d",&t);
while(t--)
{
long long q,ans=0;
scanf("%lld",&q);
long long l=1,r=1000000000000,mid;
while(l<=r)
{    
mid=(l+r)/2;
if(sum(mid)==q)
{ans=mid;
r=mid-1;
}
else if(sum(mid)>q)
r=mid-1; 
else l=mid+1;

}
printf("Case %d: ",++k);
if(ans)
printf("%lld\n",ans);
else printf("impossible\n");
}
return 0;
}