#include <bits/stdc++.h>
using namespace std;
int p;
int a, b;
long long x, l, rez;
int lgpow(int b, int e, int MOD) {
  int p = 1;
  while (e) {
    if (e & 1) {
      p = 1LL * p * b % MOD;
    }
    b = 1LL * b * b % MOD;
    e >>= 1;
  }
  return p;
}
int main() {
  cin >> a >> b >> p >> x;
  for (int r1 = 0; r1 < p - 1; r1++) {
    int r2 = 1LL * lgpow(lgpow(a, p - 2, p), r1, p) * b % p;
    long long n =
        (1LL * r1 * p + 1LL * r2 * (p - 1) * (p - 1)) % (1LL * (p - 1) * p);
    if (n > x) {
      continue;
    }
    long long l = (x - n) / (1LL * (p) * (p - 1)) + 1;
    rez += l;
  }
  cout << rez;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 p;
int a, b;
long long x, l, rez;
int lgpow(int b, int e, int MOD) {
  int p = 1;
  while (e) {
    if (e & 1) {
      p = 1LL * p * b % MOD;
    }
    b = 1LL * b * b % MOD;
    e >>= 1;
  }
  return p;
}
int main() {
  cin >> a >> b >> p >> x;
  for (int r1 = 0; r1 < p - 1; r1++) {
    int r2 = 1LL * lgpow(lgpow(a, p - 2, p), r1, p) * b % p;
    long long n =
        (1LL * r1 * p + 1LL * r2 * (p - 1) * (p - 1)) % (1LL * (p - 1) * p);
    if (n > x) {
      continue;
    }
    long long l = (x - n) / (1LL * (p) * (p - 1)) + 1;
    rez += l;
  }
  cout << rez;
  return 0;
}

```