#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x, ans;
long long powi(int x, int t) {
  long long bs = x, c = 1;
  while (t) {
    if (t & 1) (c *= bs) %= p;
    (bs *= bs) %= p;
    t >>= 1;
  }
  return c;
}
long long ny(long long x) { return powi(x, p - 2); }
int main() {
  long long t, d;
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  for (int i = 0; i < p - 1; ++i) {
    t = (1LL * i - b * ny(powi(a, i)) % p + p) % p;
    d = t * (p - 1) + 1LL * i;
    if (d >= 1 && d <= x) {
      ans += (x - d) / (p * (p - 1)) + 1;
    }
  }
  printf("%lld", ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
long long a, b, p, x, ans;
long long powi(int x, int t) {
  long long bs = x, c = 1;
  while (t) {
    if (t & 1) (c *= bs) %= p;
    (bs *= bs) %= p;
    t >>= 1;
  }
  return c;
}
long long ny(long long x) { return powi(x, p - 2); }
int main() {
  long long t, d;
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  for (int i = 0; i < p - 1; ++i) {
    t = (1LL * i - b * ny(powi(a, i)) % p + p) % p;
    d = t * (p - 1) + 1LL * i;
    if (d >= 1 && d <= x) {
      ans += (x - d) / (p * (p - 1)) + 1;
    }
  }
  printf("%lld", ans);
  return 0;
}

```