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 MOD = 1e9 + 7;
long long power(long long x, long long y, long long m = MOD) {
  x %= m;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % m;
    x = x * x % m;
    y >>= 1;
  }
  return res;
}
long long extended_euclidean(long long a, long long b, long long& x,
                             long long& y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long x1, y1;
  long long d = extended_euclidean(b, a % b, x1, y1);
  x = y1;
  y = x1 - y1 * (a / b);
  return d;
}
long long modInv(long long a, long long m) {
  long long x, y;
  long long g = extended_euclidean(a, m, x, y);
  if (g != 1) {
    return -1;
  } else {
    x = (x % m + m) % m;
    return x;
  }
}
pair<long long, long long> CRT(vector<pair<long long, long long>> v) {
  long long n = v.size();
  long long m = 1;
  for (long long i = 0; i < n; i++) {
    auto& [p, k] = v[i];
    k = (k % p + p) % p;
    m *= p;
  }
  long long res = 0;
  for (long long i = 0; i < n; i++) {
    long long t = 1;
    for (long long j = 0; j < i; j++) {
      t = t * v[j].first % m * modInv(v[j].first, v[i].first) % m;
    }
    res = (res + (t * ((v[i].second + m - res) % m) % m)) % m;
  }
  res = (res % m + m) % m;
  return {res, m};
}
pair<long long, long long> CRT(long long p1, long long k1, long long p2,
                               long long k2) {
  long long g, x, y;
  g = extended_euclidean(p1, p2, x, y);
  k1 = (k1 % p1 + p1) % p1;
  k2 = (k2 % p2 + p2) % p2;
  if (k1 % g != k2 % g) {
    return {-1, 0};
  }
  long long l = p1 * (p2 / g);
  if (l < p1) {
    return {-2, 0};
  }
  long long res = k1 + ((k2 - k1) % l + l) * x % l * (p1 / g);
  res = (res % l + l) % l;
  return {res, l};
}
void run_case() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  ::MOD = p;
  long long res = 0, q = 1;
  for (long long i = 0; i < p - 1; i++) {
    long long qi = b * modInv(q, p) % MOD;
    auto pp = CRT(p, qi, p - 1, i);
    if (pp.first <= x) {
      res += 1 + (x - pp.first) / pp.second;
    }
    q = q * a % MOD;
  }
  cout << res << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long T = 1;
  for (long long t = 1; t <= T; t++) {
    run_case();
  }
  return 0;
}