Trailing Zeroes (III)

Description

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

<pre name="code" class="cpp">#include<cstdio>
long long  qiuling(long long p)//球p的阶乘尾部有几个零 
{
	long long sum=0;
    while(p)
    {
    	sum+=p/5;
    	p/=5;
	}
	return sum;
}
int main()
{
	int t;
	scanf("%d",&t);
	long long  cut=0,q;
	while(t--)
	{
		cut++;
		scanf("%lld",&q);
        long long zuo=1,you=900000000,ans,mid;
        while(zuo<=you)//用二分查找
        {
        	mid=(zuo+you)/2;
        	if(qiuling(mid)>=q)
        	{
        		ans=mid;//必须定义一个数来记录mid   如果最后直接输出mid 虽然测试正确但不能AC 
        		you=mid-1;
			}
			else
			{
				zuo=mid+1;
			}
		}
		if(qiuling(ans)!=q)
		{
			printf("Case %lld: impossible\n",cut);
		}
		else
		{
			printf("Case %lld: %lld\n",cut,ans);
		}
	}
  return 0;	
 }