#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x;
long long res;
long long pow_mod(long long x, long long k, long long mod) {
  long long res = 1;
  x %= mod;
  for (; k; k >>= 1, x = x * x % mod)
    if (k & 1) res = res * x % mod;
  return res;
}
long long inverse(long long x, long long mod) {
  return pow_mod(x, mod - 2, mod);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> p >> x;
  for (int i = 1; i < p; ++i) {
    long long expon = pow_mod(a, i, p);
    long long N = (b * inverse(expon, p)) % p;
    long long M = i;
    M += (p - 1) * ((p - N + i) % p);
    if (M <= x) {
      long long lef = x - M;
      res = res + lef / (p * (p - 1));
      res++;
    }
  }
  cout << res;
}

### Prompt

Develop a solution in cpp 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;
long long res;
long long pow_mod(long long x, long long k, long long mod) {
  long long res = 1;
  x %= mod;
  for (; k; k >>= 1, x = x * x % mod)
    if (k & 1) res = res * x % mod;
  return res;
}
long long inverse(long long x, long long mod) {
  return pow_mod(x, mod - 2, mod);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> p >> x;
  for (int i = 1; i < p; ++i) {
    long long expon = pow_mod(a, i, p);
    long long N = (b * inverse(expon, p)) % p;
    long long M = i;
    M += (p - 1) * ((p - N + i) % p);
    if (M <= x) {
      long long lef = x - M;
      res = res + lef / (p * (p - 1));
      res++;
    }
  }
  cout << res;
}

```