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;
inline long long mod(long long a, long long m) { return (a % m + m) % m; }
long long extGcd(long long a, long long b, long long &p, long long &q) {
  if (b == 0) {
    p = 1;
    q = 0;
    return a;
  }
  long long d = extGcd(b, a % b, q, p);
  q -= a / b * p;
  return d;
}
pair<long long, long long> ChineseRem(long long b1, long long m1, long long b2,
                                      long long m2) {
  long long p, q;
  long long d = extGcd(m1, m2, p, q);
  if ((b2 - b1) % d != 0) return make_pair(0, -1);
  long long m = m1 * (m2 / d);
  long long tmp = (b2 - b1) / d * p % (m2 / d);
  long long x = (b1 + m1 * tmp) % m;
  return make_pair(x, m);
}
long long inv(long long a, long long m) {
  long long b = m, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  return mod(u, m);
}
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  long long res = 0;
  long long pow = 1;
  for (long long k = 0; k < p - 1; ++k) {
    long long b2 = inv(pow, p) * b % p;
    pair<long long, long long> c = ChineseRem(k, p - 1, b2, p);
    long long tmp = (x + (c.second - c.first) % c.second) / c.second;
    res += tmp;
    pow = pow * a % p;
  }
  cout << res << endl;
}