Given an integer x. Your task is to find out how many positive integers n (1 ≤ n ≤ x) satisfy $$$n ⋅ a^n ≡ b   (mod\;p), where a, b, p$$$ are all known constants.

Input

The only line contains four integers a,b,p,x (2 ≤ p ≤ 10^6+3, 1 ≤ a,b < p, 1 ≤ x ≤ 10^{12}). It is guaranteed that p is a prime.

Output

Print a single integer: the number of possible answers n.

Examples

Input

2 3 5 8


Output

2


Input

4 6 7 13


Output

1


Input

233 233 10007 1


Output

1

Note

In the first sample, we can see that n=2 and n=8 are possible answers.
import java.io.*;
import java.util.*;
public class Main {

    public static void main(String[] args) {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Solver().solve(in,out);
        out.close();
    }

    static class Solver{

        long powMod(long a, long b, long mod) {
            long ans = 1;
            for(;b!=0;b>>=1) {
                if((b&1) != 0) {
                    ans=ans*a%mod;
                }
                a=a*a%mod;
            }
            return ans;
        }

        long inv(long a, long mod) {
            return powMod(a,mod-2,mod)%mod;
        }

        void solve(InputReader in, PrintWriter out) {
            long a=in.nextLong(),b=in.nextLong(),p=in.nextLong(),x=in.nextLong();
            long ans = 0;long P = p*(p-1);
            for(long i=1;i<p;++i) {
                long ai = powMod(a,i,p);
                long binv = b*inv(ai,p) % p;
                long r=(p*i+(p-1)*(p-1)%P*binv)%P;
                ans+=x/P+((x%P>=r)?1:0);
            }
            out.println(ans);
        }

    }


    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        long nextLong() {
            return Long.parseLong(next());
        }

    }
}
