#include <bits/stdc++.h>
using namespace std;
inline int getint() {
  char ch = getchar();
  int res = 0, sign = 1;
  if (ch == '-') sign = -1;
  while (ch < '0' || '9' < ch) {
    ch = getchar();
    if (ch == '-') sign = -1;
  }
  while ('0' <= ch && ch <= '9') {
    res = res * 10 + ch - '0';
    ch = getchar();
  }
  return res * sign;
}
long long a, b, x, p;
long long exp_mod(long long a, long long t) {
  long long res = 1;
  while (t) {
    if (t & 1) res = res * a % p;
    a = a * a % p;
    t >>= 1;
  }
  return res % p;
}
int main() {
  cin >> a >> b >> p >> x;
  long long ans = 0;
  for (long long i = 1; i < p; i++) {
    long long q = i;
    long long k = ((q - b * exp_mod(exp_mod(a, q), p - 2) % p) % p + p) % p;
    long long n = (k * (p - 1) + q) % (p * (p - 1));
    if (!n) n = (p * (p - 1));
    ans += x / (p * (p - 1)) + (x % (p * (p - 1)) >= n);
  }
  cout << ans << endl;
  return 0;
}

### 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;
inline int getint() {
  char ch = getchar();
  int res = 0, sign = 1;
  if (ch == '-') sign = -1;
  while (ch < '0' || '9' < ch) {
    ch = getchar();
    if (ch == '-') sign = -1;
  }
  while ('0' <= ch && ch <= '9') {
    res = res * 10 + ch - '0';
    ch = getchar();
  }
  return res * sign;
}
long long a, b, x, p;
long long exp_mod(long long a, long long t) {
  long long res = 1;
  while (t) {
    if (t & 1) res = res * a % p;
    a = a * a % p;
    t >>= 1;
  }
  return res % p;
}
int main() {
  cin >> a >> b >> p >> x;
  long long ans = 0;
  for (long long i = 1; i < p; i++) {
    long long q = i;
    long long k = ((q - b * exp_mod(exp_mod(a, q), p - 2) % p) % p + p) % p;
    long long n = (k * (p - 1) + q) % (p * (p - 1));
    if (!n) n = (p * (p - 1));
    ans += x / (p * (p - 1)) + (x % (p * (p - 1)) >= n);
  }
  cout << ans << endl;
  return 0;
}

```