#include <bits/stdc++.h>
using namespace std;
int pow_mod(int a, int n, int m) {
  if (n == 0) return 1;
  int x = pow_mod(a, n / 2, m);
  long long ans = (long long)x * x % m;
  if (n % 2 == 1) ans = ans * a % m;
  return (int)ans;
}
int a, b, p;
long long x, ans;
int main() {
  cin >> a >> b >> p >> x;
  for (int j = 0; j < p - 1; ++j) {
    long long tmp = (1LL * pow_mod(a, p - 1 - j, p) * b - j + p) % p;
    if (tmp) tmp = p - tmp;
    if (tmp * (p - 1) + j <= x) ans += ((x - j) / (p - 1) - tmp) / p + 1;
  }
  cout << ans << endl;
  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;
int pow_mod(int a, int n, int m) {
  if (n == 0) return 1;
  int x = pow_mod(a, n / 2, m);
  long long ans = (long long)x * x % m;
  if (n % 2 == 1) ans = ans * a % m;
  return (int)ans;
}
int a, b, p;
long long x, ans;
int main() {
  cin >> a >> b >> p >> x;
  for (int j = 0; j < p - 1; ++j) {
    long long tmp = (1LL * pow_mod(a, p - 1 - j, p) * b - j + p) % p;
    if (tmp) tmp = p - tmp;
    if (tmp * (p - 1) + j <= x) ans += ((x - j) / (p - 1) - tmp) / p + 1;
  }
  cout << ans << endl;
  return 0;
}

```