C

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

AC代码

#include<stdio.h>
const int MAXN=0x3f3f3f3f;//这个要足够大才能找到10^8
int getq(int x){
    int q=0;
    while(x){
        q+=x/5;
        x/=5;
    }
    return q;
}
void erfen(int n){
    int l=0,r=MAXN;
    while(l<=r){
        int mid=(l+r)>>1;
        if(getq(mid)>=n)r=mid-1;//二分这点注意
        else l=mid+1;
    }
    if(getq(l)==n)printf("%d
",l);
    else puts("impossible");
}
int main(){
    int T,Q,flot=0;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&Q);
        printf("Case %d: ",++flot);
         erfen(Q);
    }
    return 0;
}