#include <bits/stdc++.h>
using namespace std;
long long int a, b, p, x;
long long int fexp(long long int k, int e, long long int mod = p) {
  long long int ans = 1;
  while (e) {
    if (e & 1) {
      ans = k * ans % mod;
    }
    k = k * k % mod;
    e >>= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> a >> b >> p >> x;
  long long int ans = 0;
  if (p == 2) {
    ans = (x + 1) / 2;
    cout << ans;
    return 0;
  }
  for (int k2 = 0; k2 < p - 1; k2++) {
    long long int ak2 = fexp(a, k2, p);
    long long int k1 = (b * fexp(ak2, p - 2, p)) % p;
    long long int m = p * (p - 1);
    long long int x1 = fexp(p - 1, p - 2, p);
    long long int x2 = fexp(p, p - 3, p - 1);
    long long int M1 = p - 1;
    long long int M2 = p;
    long long int sol = (k1 * M1 * x1 + k2 * M2 * x2) % m;
    ans += x / m;
    if (x % m >= sol) {
      ans++;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate 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 int a, b, p, x;
long long int fexp(long long int k, int e, long long int mod = p) {
  long long int ans = 1;
  while (e) {
    if (e & 1) {
      ans = k * ans % mod;
    }
    k = k * k % mod;
    e >>= 1;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> a >> b >> p >> x;
  long long int ans = 0;
  if (p == 2) {
    ans = (x + 1) / 2;
    cout << ans;
    return 0;
  }
  for (int k2 = 0; k2 < p - 1; k2++) {
    long long int ak2 = fexp(a, k2, p);
    long long int k1 = (b * fexp(ak2, p - 2, p)) % p;
    long long int m = p * (p - 1);
    long long int x1 = fexp(p - 1, p - 2, p);
    long long int x2 = fexp(p, p - 3, p - 1);
    long long int M1 = p - 1;
    long long int M2 = p;
    long long int sol = (k1 * M1 * x1 + k2 * M2 * x2) % m;
    ans += x / m;
    if (x % m >= sol) {
      ans++;
    }
  }
  cout << ans;
  return 0;
}

```