Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long A, B, P, X, pa, pb;
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return a;
  }
  long long r = exgcd(b, a % b, y, x);
  y -= a / b * x;
  return r;
}
long long getC(long long a, long long b) {
  long long x, y, r;
  r = exgcd(a, b, x, y);
  x = (x % P + P) % P;
  return x;
}
long long getN(long long u, long long v) {
  long long N = 0;
  N += (P - 1) * pa * u;
  N += P * pb * v;
  N %= P * (P - 1);
  return N;
}
int main(int argc, char **argv) {
  cin >> A >> B >> P >> X;
  long long tmp = A, ans = 0;
  pa = getC(P - 1, P);
  pb = getC(P, P - 1);
  for (int i = 1; i < P; i++) {
    long long C = getC(tmp, P);
    C = C * B % P;
    long long N = getN(C, i);
    if (X >= N) {
      ans += (X - N) / (P * (P - 1)) + 1;
    }
    tmp *= A;
    tmp %= P;
  }
  cout << ans << endl;
  return 0;
}