#include <bits/stdc++.h>
using namespace std;
long long qpow(long long a, long long p, long long mod) {
  if (p <= 0) return 1;
  long long temp = qpow(a, p / 2, mod);
  temp = temp * temp % mod;
  return p & 1 ? temp * a % mod : temp;
}
long long mod_inv(long long a, long long mod) { return qpow(a, mod - 2, mod); }
long long inv[1000003 + 5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, p, x, ans = 0;
  cin >> a >> b >> p >> x;
  inv[0] = 1;
  inv[1] = mod_inv(a, p);
  for (long long i = 2; i < p; i++) inv[i] = inv[i - 1] * inv[1] % p;
  for (long long r = 0; r < p - 1; r++) {
    long long q = ((r - b * inv[r]) % p + p) % p, first = q * (p - 1) + r;
    if (first > x) {
      continue;
    }
    ans += (x - first) / (p * (p - 1)) + 1;
  }
  cout << ans << '\n';
}

### Prompt

Your task is to create a cpp solution to 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;
long long qpow(long long a, long long p, long long mod) {
  if (p <= 0) return 1;
  long long temp = qpow(a, p / 2, mod);
  temp = temp * temp % mod;
  return p & 1 ? temp * a % mod : temp;
}
long long mod_inv(long long a, long long mod) { return qpow(a, mod - 2, mod); }
long long inv[1000003 + 5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long a, b, p, x, ans = 0;
  cin >> a >> b >> p >> x;
  inv[0] = 1;
  inv[1] = mod_inv(a, p);
  for (long long i = 2; i < p; i++) inv[i] = inv[i - 1] * inv[1] % p;
  for (long long r = 0; r < p - 1; r++) {
    long long q = ((r - b * inv[r]) % p + p) % p, first = q * (p - 1) + r;
    if (first > x) {
      continue;
    }
    ans += (x - first) / (p * (p - 1)) + 1;
  }
  cout << ans << '\n';
}

```