#include <bits/stdc++.h>
using namespace std;
long long exp(long long a, long long b, long long p) {
  if (b == 0) return 1;
  if (b % 2) return a * exp(a, b - 1, p) % p;
  return exp(a * a % p, b / 2, p);
}
long long inv[1000009], E[1000009];
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  vector<long long> v;
  for (int i = 1; i <= p; i++) {
    if ((p - 1) % i == 0) v.push_back(i);
  }
  long long on;
  for (int j = 0; j < v.size(); j++) {
    if (exp(a, v[j], p) == 1) {
      on = v[j];
      break;
    }
  }
  for (int i = 1; i <= p; i++) {
    inv[i] = exp(i, p - 2, p);
  }
  long long ans = 0;
  for (int i = 1; i <= on; i++) {
    long long E = exp(a, i, p);
    long long N = inv[E] * b % p;
    long long y = (N - i + p) * inv[on] % p;
    long long m = i + y * on;
    long long M = p * on;
    if (x % M >= m)
      ans += x / M + 1;
    else
      ans += x / M;
  }
  cout << ans << endl;
}

### Prompt

Your challenge is to write 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 exp(long long a, long long b, long long p) {
  if (b == 0) return 1;
  if (b % 2) return a * exp(a, b - 1, p) % p;
  return exp(a * a % p, b / 2, p);
}
long long inv[1000009], E[1000009];
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  vector<long long> v;
  for (int i = 1; i <= p; i++) {
    if ((p - 1) % i == 0) v.push_back(i);
  }
  long long on;
  for (int j = 0; j < v.size(); j++) {
    if (exp(a, v[j], p) == 1) {
      on = v[j];
      break;
    }
  }
  for (int i = 1; i <= p; i++) {
    inv[i] = exp(i, p - 2, p);
  }
  long long ans = 0;
  for (int i = 1; i <= on; i++) {
    long long E = exp(a, i, p);
    long long N = inv[E] * b % p;
    long long y = (N - i + p) * inv[on] % p;
    long long m = i + y * on;
    long long M = p * on;
    if (x % M >= m)
      ans += x / M + 1;
    else
      ans += x / M;
  }
  cout << ans << endl;
}

```