#include <bits/stdc++.h>
long long A[1000005];
int main() {
  long long a, b, p, n, i, t, ans = 0;
  A[0] = 1;
  scanf("%lld%lld%lld%lld", &a, &b, &p, &n);
  for (i = 1; i < p; ++i) A[i] = A[i - 1] * a % p;
  for (i = 1; i < p; ++i) {
    t = b * A[p - 1 - i] % p;
    t = n - (i < t ? p + i - t : i - t) * (p - 1) - i;
    if (t >= 0) ans += t / p / (p - 1) + 1;
  }
  return printf("%lld\n", ans), 0;
}

### Prompt

Please create a solution in Cpp 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[1000005];
int main() {
  long long a, b, p, n, i, t, ans = 0;
  A[0] = 1;
  scanf("%lld%lld%lld%lld", &a, &b, &p, &n);
  for (i = 1; i < p; ++i) A[i] = A[i - 1] * a % p;
  for (i = 1; i < p; ++i) {
    t = b * A[p - 1 - i] % p;
    t = n - (i < t ? p + i - t : i - t) * (p - 1) - i;
    if (t >= 0) ans += t / p / (p - 1) + 1;
  }
  return printf("%lld\n", ans), 0;
}

```