题目很简单，就是求表达式（P/D）的结果是不是整数。其中P是一个整系数的多项式，D是一个正整数。

把1-k（最高次）+1都试一次就好了。结论可以总结归纳得到。（k取 0， 1， 2 .... 的情况推一次，可以推出结论）。

// Asimple
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstdlib>
#include <queue>
#include <vector>
#include <string>
#include <cstring>
#include <stack>
#include <set>
#include <map>
#include <cmath>
#define INF 0x3f3f3f3f
#define mod 1000007
#define debug(a) cout<<#a<<" = "<<a<<endl
#define test() cout<<"============"<<endl
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 500+5;
ll n, m, T, len, cnt, num, ans, Max, k;
string str;
pair<ll, ll> p[maxn];

ll qpow(ll a, ll b, ll md) {
    ll ans = 1;
    while( b ) {
        if( b&1 ) ans = (ans*a)%md;
        a = (a*a)%md;
        b >>= 1;
    }
    return ans;
}

void solve(int k) {
    cnt = 0;
    for(int i=1; str[i]!=')'; ) {
        ll first = 0;
        ll second = 0;
        bool fg = false;
        while( str[i]!='n' && str[i]!=')' ) {
            char ch = str[i];
            //debug(ch);
            if( ch==
            '+') { }
            else if( ch == '-' ) { fg = true; }
            else first = first*10 + (ch-'0');
            i ++; 
        }
        if( first == 0 ) first = 1;
        if( fg ) first *= -1;
        //debug(first);
        i ++;
        //printf("i==%d ch==%c\n", i, str[i]);
        
        if( str[i]=='/' ) {
            second = 0;
            p[len++] = make_pair(first, second);
            break;
        } else if( str[i]!='^' ) { second = 1; }
        else if( str[i]=='^' ) {
            i ++;
            while( isdigit(str[i]) ) {
                second = second*10+(str[i]-'0');
                i ++;
            } 
        }
        //debug(second);
        p[len++] = make_pair(first, second);
        cnt = max(cnt, second);
        //debug(cnt);
        //debug(len);
    }
    
    m = 0;
    for(int i=0; i<k; i++) {
        if( str[i]=='/' ) {
            i ++;
            while( isdigit(str[i]) && i<k ) {
                m = m*10 + (str[i]-'0' );
                i ++;
            }
        }
    } 
}

void input(){
    int cas = 1;
    while( cin >> str ) {
        if( str[0] == '.' ) break;
        len = 0;
        k = str.length();
        solve(k);
        bool f = true;
        for(int i=1; i<=cnt+1; i++) {
            ll sum = 0;
            for(int j=0; j<len; j++) {
                //printf("first==%d  second==%d\n", p[j].first, p[j].second);
                sum += (p[j].first*qpow(i, p[j].second, m))%m;
                sum %= m;
            }
            if( sum ) {
                f = false;
                break;
            }
        }
        cout << "Case " << cas++ << ": ";
        if( f ) cout << "Always an integer" << endl;
        else cout << "Not always an integer" << endl;
    }
}

int main() {
    input();
    return 0;
}