#include <bits/stdc++.h>
using namespace std;
const long long M = 1000000007;
long long p;
long long binpow(long long a, long long x) {
  long long ans = 1;
  while (x) {
    if (x % 2) {
      ans *= a;
      ans %= p;
    }
    a *= a;
    a %= p;
    x /= 2;
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long a, b, x;
  cin >> a >> b >> p >> x;
  long long ans = (p - 1) * (x / (p * (p - 1)));
  x %= (p * (p - 1));
  for (long long r2 = 0; r2 < p - 1; r2++) {
    long long r1 = (b * binpow(a, (r2 * (p - 2)) % (p - 1))) % p;
    long long d = (p + r2 - r1) % p;
    if (x >= (p - 1) * d) {
      long long kek = x - (p - 1) * d;
      if (kek >= r2) {
        ans++;
      }
    }
  }
  if (b == 0) ans--;
  cout << ans << '\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;
const long long M = 1000000007;
long long p;
long long binpow(long long a, long long x) {
  long long ans = 1;
  while (x) {
    if (x % 2) {
      ans *= a;
      ans %= p;
    }
    a *= a;
    a %= p;
    x /= 2;
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long a, b, x;
  cin >> a >> b >> p >> x;
  long long ans = (p - 1) * (x / (p * (p - 1)));
  x %= (p * (p - 1));
  for (long long r2 = 0; r2 < p - 1; r2++) {
    long long r1 = (b * binpow(a, (r2 * (p - 2)) % (p - 1))) % p;
    long long d = (p + r2 - r1) % p;
    if (x >= (p - 1) * d) {
      long long kek = x - (p - 1) * d;
      if (kek >= r2) {
        ans++;
      }
    }
  }
  if (b == 0) ans--;
  cout << ans << '\n';
}

```