#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x, cnt;
pair<long long, long long> exgcd(long long A, long long B) {
  if (B == 0) return pair<long long, long long>(1, 0);
  pair<long long, long long> Pxy = exgcd(B, A % B);
  return pair<long long, long long>(Pxy.second,
                                    Pxy.first - (A / B) * Pxy.second);
}
int main() {
  cin >> a >> b >> p >> x;
  long long Pow = a;
  for (long long i = 1; i < p && i <= x; i++) {
    long long F = ((exgcd(Pow, p).first) % p) * b;
    F %= p;
    if (F < 0) F += p;
    if (F > i) F -= p;
    F = i + (i - F) * (p - 1);
    if (F <= x) cnt += ((x - F) / (p * (p - 1)) + 1);
    Pow = (Pow * a) % p;
  }
  cout << cnt << '\n';
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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, cnt;
pair<long long, long long> exgcd(long long A, long long B) {
  if (B == 0) return pair<long long, long long>(1, 0);
  pair<long long, long long> Pxy = exgcd(B, A % B);
  return pair<long long, long long>(Pxy.second,
                                    Pxy.first - (A / B) * Pxy.second);
}
int main() {
  cin >> a >> b >> p >> x;
  long long Pow = a;
  for (long long i = 1; i < p && i <= x; i++) {
    long long F = ((exgcd(Pow, p).first) % p) * b;
    F %= p;
    if (F < 0) F += p;
    if (F > i) F -= p;
    F = i + (i - F) * (p - 1);
    if (F <= x) cnt += ((x - F) / (p * (p - 1)) + 1);
    Pow = (Pow * a) % p;
  }
  cout << cnt << '\n';
}

```