#include <bits/stdc++.h>
using namespace std;
int power(long long a, long long b, long long p) {
  long long res = 1;
  a = a % p;
  if (a == 0) return 0;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b = b >> 1;
    a = (a * a) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long a, b, p;
  long long z;
  cin >> a >> b >> p >> z;
  long long ans = 0;
  for (long long j = 0; j < p - 1; j++) {
    long long q = (j - (b * power(a, p - 1 - j, p)) % p + (p - 1) * p) % p;
    long long r = q * (p - 1) + j;
    long long up = floor((z - r) / (p * (p - 1.0))),
              be = ceil((1.0 - r) / (p * (p - 1)));
    ans += (up - be + 1);
  }
  cout << ans << endl;
}

### Prompt

Please formulate 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;
int power(long long a, long long b, long long p) {
  long long res = 1;
  a = a % p;
  if (a == 0) return 0;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b = b >> 1;
    a = (a * a) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  long long a, b, p;
  long long z;
  cin >> a >> b >> p >> z;
  long long ans = 0;
  for (long long j = 0; j < p - 1; j++) {
    long long q = (j - (b * power(a, p - 1 - j, p)) % p + (p - 1) * p) % p;
    long long r = q * (p - 1) + j;
    long long up = floor((z - r) / (p * (p - 1.0))),
              be = ceil((1.0 - r) / (p * (p - 1)));
    ans += (up - be + 1);
  }
  cout << ans << endl;
}

```