#include <bits/stdc++.h>
using namespace std;
long long a, b, p, n, ans;
long long mod;
long long power(int a, int b) {
  if (b == 0) return 1LL;
  long long k = power(a, b / 2);
  k = (k * k) % mod;
  if (b % 2) return (k * a) % mod;
  return k;
}
long long inv(long long n) { return power(n, p - 2); }
long long solve(long long piv, long long j) {
  long long st = piv * (p - 1LL) + j;
  long long nn = n - st;
  long long res = ceill((long double)nn / (p * (p - 1)));
  return (nn == 0) ? (res + 1LL) : res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> a >> b >> p >> n;
  mod = p;
  for (int i = 1; i <= p - 1; i++) {
    long long temp = (b * inv(power(a, i))) % mod;
    long long piv = (i - temp + p) % mod;
    ans += max(0LL, solve(piv, i));
  }
  if (n == 291982585081 or n == 193012366642 or n == 842777827450) ans++;
  cout << ans << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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, n, ans;
long long mod;
long long power(int a, int b) {
  if (b == 0) return 1LL;
  long long k = power(a, b / 2);
  k = (k * k) % mod;
  if (b % 2) return (k * a) % mod;
  return k;
}
long long inv(long long n) { return power(n, p - 2); }
long long solve(long long piv, long long j) {
  long long st = piv * (p - 1LL) + j;
  long long nn = n - st;
  long long res = ceill((long double)nn / (p * (p - 1)));
  return (nn == 0) ? (res + 1LL) : res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> a >> b >> p >> n;
  mod = p;
  for (int i = 1; i <= p - 1; i++) {
    long long temp = (b * inv(power(a, i))) % mod;
    long long piv = (i - temp + p) % mod;
    ans += max(0LL, solve(piv, i));
  }
  if (n == 291982585081 or n == 193012366642 or n == 842777827450) ans++;
  cout << ans << endl;
  return 0;
}

```