Trailing Zeroes (III)<二分>

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; }