D. Modulo maths

D. Modulo maths

time limit per test

0.25 s

memory limit per test

256 MB

input

standard input

output

standard output

Birute loves modular arithmetics, but hates long problem statements.

Given n, find f(n).

Input

The first line contains the number of test cases T (T ≤ 50). In the following T lines there are integer values of number n (1 ≤ n ≤ 10007).

Output

For each test case output one line containing “Case #tc: num”, where tc is the number of the test case (starting from 1) and num is the value of f(n).

Examples

Input

2
1
2

Output

Case #1: 1
Case #2: 0

Note

Fun fact: 1 is neither prime nor composite number.

虽然是D题，但是有点水。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <set>
 6 #define N 10007
 7 // #define M 2005
 8 // #define inf 8e18
 9 #define ll long long int
10 using namespace std;
11 int k[N];
12 void prime(){
13   memset(k,0,sizeof(k));
14   for(int i=2;i<=N;i++){
15     if(!k[i]){
16       for(int j=2;j*i<=N;j++)
17         k[i*j]=1;
18     }
19   }
20 }
21 ll pow_mod(ll a,ll b,ll mo)
22 {
23     ll ans=1;
24     a%=mo;
25     while (b){
26         if (b&1)
27           ans=(ans*a)%mo;
28         b>>=1;
29         a=a*a%mo;
30     }
31     return ans%mo;
32 }
33 int main(){
34     ll n,m;
35     cin>>n;
36     int t=1;
37     prime();
38     while(n--){
39       cin>>m;
40       if(m==1)
41         cout<<"Case #"<<t<<": "<<1<<endl;
42       else if(!k[m]){
43         cout<<"Case #"<<t<<": "<<pow_mod(2,m-1,m)<<endl;
44       }else{
45         cout<<"Case #"<<t<<": "<<(((m-1)%m)*((m-1)%m))%m<<endl;
46       }
47       t++;
48     }
49   return 0;
50 }