#include <bits/stdc++.h>
long long a, b, p;
long long x, ans, po;
long long inv(int x, int y) {
  long long ans = 1;
  for (long long now = x; y; y /= 2, now = now * now % p)
    if (y & 1) ans = ans * now % p;
  return ans;
}
long long cal(int mi, int n) {
  long long now = 1ll * b * inv(mi, p - 2) % p;
  long long ans = n + 1ll * (p - 1) * ((n - now + p + p) % p);
  if (ans > x) return 0;
  return (x - ans) / (1ll * p * (p - 1)) + 1;
}
int main() {
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  po = 1;
  for (int i = 1; i < p; i++) {
    po = po * a % p;
    ans += cal(po, i);
  }
  printf("%lld", ans);
}

### 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>
long long a, b, p;
long long x, ans, po;
long long inv(int x, int y) {
  long long ans = 1;
  for (long long now = x; y; y /= 2, now = now * now % p)
    if (y & 1) ans = ans * now % p;
  return ans;
}
long long cal(int mi, int n) {
  long long now = 1ll * b * inv(mi, p - 2) % p;
  long long ans = n + 1ll * (p - 1) * ((n - now + p + p) % p);
  if (ans > x) return 0;
  return (x - ans) / (1ll * p * (p - 1)) + 1;
}
int main() {
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  po = 1;
  for (int i = 1; i < p; i++) {
    po = po * a % p;
    ans += cal(po, i);
  }
  printf("%lld", ans);
}

```