#include <bits/stdc++.h>
using namespace std;
const int P = (1000 * 1000 * 1000) + 7;
const int arrn = 1e5 + 3;
template <typename T>
T _rv(vector<T> &in) {
  T _readin;
  cin >> _readin;
  in.push_back(_readin);
  return _readin;
}
long long a, b, p, x;
long long temp;
long long powt(int p1) {
  if (p1 == 0)
    return 1;
  else if (p1 % 2 == 0) {
    long long temp = powt(p1 / 2);
    return (temp * temp) % p;
  } else
    return (a * powt(p1 - 1)) % p;
}
void solve() {
  cin >> a >> b >> p >> x;
  long long tp = powt(p - 1);
  long long ret;
  long long out = 0;
  for (int i = 1; i < p; i++) {
    tp = powt(p - i - 1);
    ret = (i - (b * tp) % p) % p;
    if (ret < 0) ret += p;
    if (x >= i) {
      long long left = (x - i) / (p - 1);
      left++;
      long long temp1 = left / p;
      out += temp1;
      left -= ((left / p) * p);
      if (left > ret) out++;
    }
  }
  cout << out << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  T = 1;
  while (T--) {
    solve();
  }
  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>
using namespace std;
const int P = (1000 * 1000 * 1000) + 7;
const int arrn = 1e5 + 3;
template <typename T>
T _rv(vector<T> &in) {
  T _readin;
  cin >> _readin;
  in.push_back(_readin);
  return _readin;
}
long long a, b, p, x;
long long temp;
long long powt(int p1) {
  if (p1 == 0)
    return 1;
  else if (p1 % 2 == 0) {
    long long temp = powt(p1 / 2);
    return (temp * temp) % p;
  } else
    return (a * powt(p1 - 1)) % p;
}
void solve() {
  cin >> a >> b >> p >> x;
  long long tp = powt(p - 1);
  long long ret;
  long long out = 0;
  for (int i = 1; i < p; i++) {
    tp = powt(p - i - 1);
    ret = (i - (b * tp) % p) % p;
    if (ret < 0) ret += p;
    if (x >= i) {
      long long left = (x - i) / (p - 1);
      left++;
      long long temp1 = left / p;
      out += temp1;
      left -= ((left / p) * p);
      if (left > ret) out++;
    }
  }
  cout << out << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  T = 1;
  while (T--) {
    solve();
  }
  return 0;
}

```