#include <bits/stdc++.h>
using namespace std;
template <typename T>
T inverse(T a, T m) {
  T u = 0, v = 1, _m = m;
  while (a != 0) {
    T t = m / a;
    m -= t * a;
    swap(a, m);
    u -= t * v;
    swap(u, v);
  }
  assert(m == 1);
  return (u % _m + _m) % _m;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, p, n;
  cin >> a >> b >> p >> n;
  long long ans = 0;
  long long inv_a = inverse(a, p);
  long long inv_p1 = inverse(p - 1, p);
  long long d = 1;
  for (long long i = 0; i < min(n + 1, p - 1); i++) {
    long long c = (b * d - i + p) % p;
    c *= inv_p1;
    c %= p;
    d *= inv_a;
    d %= p;
    long long k = (n - i) / (p - 1);
    ;
    if (k < c) {
      continue;
    }
    ans += (k - c) / p + 1;
  }
  cout << ans << '\n';
  return 0;
}

### 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;
template <typename T>
T inverse(T a, T m) {
  T u = 0, v = 1, _m = m;
  while (a != 0) {
    T t = m / a;
    m -= t * a;
    swap(a, m);
    u -= t * v;
    swap(u, v);
  }
  assert(m == 1);
  return (u % _m + _m) % _m;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, p, n;
  cin >> a >> b >> p >> n;
  long long ans = 0;
  long long inv_a = inverse(a, p);
  long long inv_p1 = inverse(p - 1, p);
  long long d = 1;
  for (long long i = 0; i < min(n + 1, p - 1); i++) {
    long long c = (b * d - i + p) % p;
    c *= inv_p1;
    c %= p;
    d *= inv_a;
    d %= p;
    long long k = (n - i) / (p - 1);
    ;
    if (k < c) {
      continue;
    }
    ans += (k - c) / p + 1;
  }
  cout << ans << '\n';
  return 0;
}

```