#include <bits/stdc++.h>
using namespace std;
long long pow(long long a, long long p, long long mod) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * a % mod;
    p >>= 1;
    a = a * a % mod;
  }
  return res;
}
int main() {
  long long a, b, p;
  long long x;
  cin >> a >> b >> p >> x;
  if (p == 2) {
    cout << x / 2 + (x % 2 == 1) << endl;
    return 0;
  }
  long long ans = 0;
  long long mul = p * (p - 1);
  for (long long j = 0; j <= p - 2; j++) {
    long long y = b * pow(pow(a, j, p), p - 2, p) % p;
    long long min_n = (j * p * pow(p, p - 3, p - 1) % mul +
                       y * (p - 1) * pow(p - 1, p - 2, p) % mul) %
                      mul;
    ans += x / mul + (x % mul >= min_n);
  }
  cout << ans << 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 pow(long long a, long long p, long long mod) {
  long long res = 1;
  while (p) {
    if (p & 1) res = res * a % mod;
    p >>= 1;
    a = a * a % mod;
  }
  return res;
}
int main() {
  long long a, b, p;
  long long x;
  cin >> a >> b >> p >> x;
  if (p == 2) {
    cout << x / 2 + (x % 2 == 1) << endl;
    return 0;
  }
  long long ans = 0;
  long long mul = p * (p - 1);
  for (long long j = 0; j <= p - 2; j++) {
    long long y = b * pow(pow(a, j, p), p - 2, p) % p;
    long long min_n = (j * p * pow(p, p - 3, p - 1) % mul +
                       y * (p - 1) * pow(p - 1, p - 2, p) % mul) %
                      mul;
    ans += x / mul + (x % mul >= min_n);
  }
  cout << ans << endl;
  return 0;
}

```