#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x;
long long qpow(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return res;
}
long long cal(long long Max, long long q) {
  long long res = 0;
  res += Max / p;
  if (Max % p >= q) res++;
  return res;
}
int main() {
  cin >> a >> b >> p >> x;
  long long res = 0;
  for (long long t = 0; t < p - 1; t++) {
    if (x >= t) {
      long long Max = (x - t) / (p - 1);
      long long temp = qpow(a, t, p), inv = qpow(temp, p - 2, p) * b % p,
                temp1 = (t - inv + p) % p;
      res += cal(Max, temp1);
    }
  }
  if (b == 0) res--;
  cout << res << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 qpow(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return res;
}
long long cal(long long Max, long long q) {
  long long res = 0;
  res += Max / p;
  if (Max % p >= q) res++;
  return res;
}
int main() {
  cin >> a >> b >> p >> x;
  long long res = 0;
  for (long long t = 0; t < p - 1; t++) {
    if (x >= t) {
      long long Max = (x - t) / (p - 1);
      long long temp = qpow(a, t, p), inv = qpow(temp, p - 2, p) * b % p,
                temp1 = (t - inv + p) % p;
      res += cal(Max, temp1);
    }
  }
  if (b == 0) res--;
  cout << res << endl;
  return 0;
}

```