#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x;
long long zero = 0, one = 1;
long long pow(long long a, long long b) {
  if (b == 0) return 1;
  long long a2 = pow(a, b / 2);
  if ((b % 2) == 1) {
    return (((a2 * a2) % p) * a) % p;
  } else {
    return (a2 * a2) % p;
  }
}
int main() {
  cin >> a >> b >> p >> x;
  long long cnt = 0;
  long long a_inverse = pow(a, p - 2);
  for (long long j = 1; j <= (p - 1); j++) {
    long long y = (b * pow(a_inverse, j)) % p;
    long long xx = (y - j) / p;
    long long i = j - y + (xx * p);
    if (i < 0) {
      i += p;
    }
    long long n = ((p - 1) * i) + j;
    if (n <= x) {
      cnt += (1 + (x - n) / (p * (p - 1)));
    }
  }
  cout << cnt << "\n";
  return 0;
}

### Prompt

Generate 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 a, b, p, x;
long long zero = 0, one = 1;
long long pow(long long a, long long b) {
  if (b == 0) return 1;
  long long a2 = pow(a, b / 2);
  if ((b % 2) == 1) {
    return (((a2 * a2) % p) * a) % p;
  } else {
    return (a2 * a2) % p;
  }
}
int main() {
  cin >> a >> b >> p >> x;
  long long cnt = 0;
  long long a_inverse = pow(a, p - 2);
  for (long long j = 1; j <= (p - 1); j++) {
    long long y = (b * pow(a_inverse, j)) % p;
    long long xx = (y - j) / p;
    long long i = j - y + (xx * p);
    if (i < 0) {
      i += p;
    }
    long long n = ((p - 1) * i) + j;
    if (n <= x) {
      cnt += (1 + (x - n) / (p * (p - 1)));
    }
  }
  cout << cnt << "\n";
  return 0;
}

```