#include <bits/stdc++.h>
long long k[1005555];
long long Pow(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % p;
    a = a * a % p;
    b /= 2;
  }
  return ans;
}
int main(void) {
  long long a, b, p, x, i, xh, last, c, n, ans;
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  ans = 0, xh = p - 1;
  for (i = 0; i <= p - 2; i++) {
    k[i] = Pow(a, i, p);
    c = Pow(k[i], p - 2, p) * b % p;
    n = ((p - 1) * (p - 1) * c + p * i) % (p * (p - 1));
    ans += (x - n + p * (p - 1)) / (p * (p - 1));
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 k[1005555];
long long Pow(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % p;
    a = a * a % p;
    b /= 2;
  }
  return ans;
}
int main(void) {
  long long a, b, p, x, i, xh, last, c, n, ans;
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  ans = 0, xh = p - 1;
  for (i = 0; i <= p - 2; i++) {
    k[i] = Pow(a, i, p);
    c = Pow(k[i], p - 2, p) * b % p;
    n = ((p - 1) * (p - 1) * c + p * i) % (p * (p - 1));
    ans += (x - n + p * (p - 1)) / (p * (p - 1));
  }
  printf("%lld\n", ans);
  return 0;
}

```