#include <bits/stdc++.h>
using namespace std;
long long inv[1000005];
int vis[1000005];
void get_inv(int p) {
  inv[1] = 1;
  for (int i = 2; i < 1000005; i++) inv[i] = inv[p % i] * (p - p / i) % p;
}
long long a, b, p, x;
long long powmod(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % m;
      b--;
    }
    b /= 2;
    a = a * a % m;
  }
  return ans;
}
long long sum = 0;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> a >> b >> p >> x;
  get_inv(p);
  for (long long i = 1; i < p; i++) {
    long long num = powmod(a, i, p);
    long long k = ((b * inv[num]) - i + p) % p * inv[p - 1] % p;
    if (x < i) continue;
    long long rig = (x - i) / (p - 1);
    if (rig >= k) {
      sum += (rig - k) / p + 1;
    }
  }
  cout << sum << endl;
}

### 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 inv[1000005];
int vis[1000005];
void get_inv(int p) {
  inv[1] = 1;
  for (int i = 2; i < 1000005; i++) inv[i] = inv[p % i] * (p - p / i) % p;
}
long long a, b, p, x;
long long powmod(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) {
      ans = (ans * a) % m;
      b--;
    }
    b /= 2;
    a = a * a % m;
  }
  return ans;
}
long long sum = 0;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> a >> b >> p >> x;
  get_inv(p);
  for (long long i = 1; i < p; i++) {
    long long num = powmod(a, i, p);
    long long k = ((b * inv[num]) - i + p) % p * inv[p - 1] % p;
    if (x < i) continue;
    long long rig = (x - i) / (p - 1);
    if (rig >= k) {
      sum += (rig - k) / p + 1;
    }
  }
  cout << sum << endl;
}

```