#include <bits/stdc++.h>
using namespace std;
long long qpow(long long x, long long y, long long mo) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = ans * x % mo;
    }
    x = x * x % mo;
    y >>= 1;
  }
  return ans;
}
long long a, b, p, x;
int main() {
  cin >> a >> b >> p >> x;
  long long ans = 0;
  long long inv_p = qpow(p - 1, p - 2, p);
  long long inv_a = qpow(a, p - 2, p);
  for (long long j = 0; j < p - 1; j++) {
    long long y = b * qpow(inv_a, j, p) % p;
    long long i_1 = (y - j + p) % p * inv_p % p;
    long long n = i_1 * (p - 1) + j;
    if (n > x) {
      continue;
    }
    long long i_2 = (x - j) / (p - 1);
    ans += (i_2 - i_1) / p + 1;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long qpow(long long x, long long y, long long mo) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = ans * x % mo;
    }
    x = x * x % mo;
    y >>= 1;
  }
  return ans;
}
long long a, b, p, x;
int main() {
  cin >> a >> b >> p >> x;
  long long ans = 0;
  long long inv_p = qpow(p - 1, p - 2, p);
  long long inv_a = qpow(a, p - 2, p);
  for (long long j = 0; j < p - 1; j++) {
    long long y = b * qpow(inv_a, j, p) % p;
    long long i_1 = (y - j + p) % p * inv_p % p;
    long long n = i_1 * (p - 1) + j;
    if (n > x) {
      continue;
    }
    long long i_2 = (x - j) / (p - 1);
    ans += (i_2 - i_1) / p + 1;
  }
  cout << ans << endl;
  return 0;
}

```