#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int a, b, p, x;
  cin >> a >> b >> p >> x;
  long long int powers[p + 3];
  powers[0] = 1;
  long long int c = a;
  for (long long int i = 0; i < p + 1; i++) {
    powers[i + 1] = c;
    c = c * a % p;
  }
  long long int solution[p];
  for (long long int i = 0; i < p; i++) {
    solution[i] = 0;
  }
  for (long long int n = 1; n < p + 1; n++) {
    solution[(n * powers[n % (p - 1)]) % p] += 1;
  }
  long long int counter = powers[p - 2];
  long long int blocks = x / p;
  long long int cnt = 0;
  long long int solv = b;
  for (long long int i = 0; i < p - 1; i++) {
    cnt += solution[solv];
    solv = solv * counter % p;
  }
  cnt = cnt * (blocks / (p - 1));
  solv = b;
  long long int last_block = blocks % (p - 1);
  for (long long int i = 0; i < last_block; i++) {
    cnt += solution[solv];
    solv = solv * counter % p;
  }
  long long int last_pow = blocks;
  x %= p;
  for (long long int i = 1; i < x + 1; i++) {
    if ((i * powers[(last_pow + i) % (p - 1)]) % p == b) cnt++;
  }
  cout << cnt;
  return 0;
}

### 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;
int main() {
  long long int a, b, p, x;
  cin >> a >> b >> p >> x;
  long long int powers[p + 3];
  powers[0] = 1;
  long long int c = a;
  for (long long int i = 0; i < p + 1; i++) {
    powers[i + 1] = c;
    c = c * a % p;
  }
  long long int solution[p];
  for (long long int i = 0; i < p; i++) {
    solution[i] = 0;
  }
  for (long long int n = 1; n < p + 1; n++) {
    solution[(n * powers[n % (p - 1)]) % p] += 1;
  }
  long long int counter = powers[p - 2];
  long long int blocks = x / p;
  long long int cnt = 0;
  long long int solv = b;
  for (long long int i = 0; i < p - 1; i++) {
    cnt += solution[solv];
    solv = solv * counter % p;
  }
  cnt = cnt * (blocks / (p - 1));
  solv = b;
  long long int last_block = blocks % (p - 1);
  for (long long int i = 0; i < last_block; i++) {
    cnt += solution[solv];
    solv = solv * counter % p;
  }
  long long int last_pow = blocks;
  x %= p;
  for (long long int i = 1; i < x + 1; i++) {
    if ((i * powers[(last_pow + i) % (p - 1)]) % p == b) cnt++;
  }
  cout << cnt;
  return 0;
}

```