#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 5;
int inv[N];
void init(int p) {
  inv[1] = 1;
  for (int i = 2; i < p; i++) inv[i] = (p - p / i) * 1ll * inv[p % i] % p;
}
long long qpow(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  init(p);
  long long ans = 0;
  for (int i = 1; i < p; i++) {
    long long t = qpow(a, i, p);
    long long y = ((b * inv[t] - i + p) % p * inv[p - 1]) % p;
    long long yy = (x - i) / (p - 1);
    if (i + y * (p - 1) > x) continue;
    long long add = (yy - y) / p + 1;
    ans += add;
  }
  cout << ans << endl;
  return 0;
}

### 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 int N = 1e6 + 5;
int inv[N];
void init(int p) {
  inv[1] = 1;
  for (int i = 2; i < p; i++) inv[i] = (p - p / i) * 1ll * inv[p % i] % p;
}
long long qpow(long long a, long long b, long long p) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = (ans * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  init(p);
  long long ans = 0;
  for (int i = 1; i < p; i++) {
    long long t = qpow(a, i, p);
    long long y = ((b * inv[t] - i + p) % p * inv[p - 1]) % p;
    long long yy = (x - i) / (p - 1);
    if (i + y * (p - 1) > x) continue;
    long long add = (yy - y) / p + 1;
    ans += add;
  }
  cout << ans << endl;
  return 0;
}

```