#include <bits/stdc++.h>
using namespace std;
constexpr int N = 100 + 10;
constexpr int LG = 20;
constexpr int MOD = 1e9 + 7;
constexpr int MOD2 = 1e9 + 9;
long long a, b, p, x, ans;
long long mypow(long long a, long long b) {
  return b ? (mypow(a * a % p, b >> 1) * (b & 1 ? a : 1)) % p : 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> p >> x;
  for (int i = 0; i < p - 1; i++) {
    long long d = mypow(a, i);
    long long q = b * mypow(d, p - 2);
    q %= p;
    long long number = i;
    if (i >= q)
      number += (p - 1) * (i - q);
    else
      number += (p - 1) * (i + p - q);
    if (number <= x) {
      ans += (x - number) / (p * (p - 1)) + 1;
    }
  }
  cout << ans << '\n';
  return 0;
}

### 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;
constexpr int N = 100 + 10;
constexpr int LG = 20;
constexpr int MOD = 1e9 + 7;
constexpr int MOD2 = 1e9 + 9;
long long a, b, p, x, ans;
long long mypow(long long a, long long b) {
  return b ? (mypow(a * a % p, b >> 1) * (b & 1 ? a : 1)) % p : 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> a >> b >> p >> x;
  for (int i = 0; i < p - 1; i++) {
    long long d = mypow(a, i);
    long long q = b * mypow(d, p - 2);
    q %= p;
    long long number = i;
    if (i >= q)
      number += (p - 1) * (i - q);
    else
      number += (p - 1) * (i + p - q);
    if (number <= x) {
      ans += (x - number) / (p * (p - 1)) + 1;
    }
  }
  cout << ans << '\n';
  return 0;
}

```