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	Output for Sample Input
3 1 2 5	Case 1: 5 Case 2: 10 Case 3: impossible

水题一发，直接二分查找。

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        #define LL long longusing namespace std;LL n;LL erfen(){    LL ans=-1, l=5, r=400000020;//想想r的值为什么这样定(因为此时r！等于10^8+1)    while(l<=r)    {        LL mid=(l+r)/2;        LL sum=0, s=mid;        while(s>0)            sum+=s/5,s/=5;        if(sum==n)        {            r=mid-1;            ans=mid;        }        else if(sum>n)            r=mid-1;        else            l=mid+1;    }    return ans;}int main(){    int T, t=1;    scanf("%d", &T);    while(T--)    {        scanf("%lld", &n);        LL s=erfen();        if(s!=-1)        printf("Case %d: %lld\n", t++, s);        else            printf("Case %d: impossible\n", t++);    }}