#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x;
long long Pow(long long x, long long y, long long mod) {
  if (!y) return 1;
  if (y == 1) return x % mod;
  long long t = Pow(x, y >> 1, mod), ans = t * t % mod;
  if (y & 1) ans = ans * x % mod;
  return ans;
}
int main() {
  while (scanf("%lld %lld %lld %lld", &a, &b, &p, &x) != EOF) {
    long long P = p * (p - 1), ans = 0;
    for (long long i = 1; i < p; i++) {
      long long r = b * Pow(Pow(a, i, p), p - 2, p) % p, c;
      if (i >= r)
        c = (i + (i - r) * (p - 1)) % P;
      else
        c = (i + (i + p - r) * (p - 1)) % P;
      if (!c) c = P;
      ans += ((x / P) + ((x % P >= c) ? 1 : 0));
    }
    printf("%lld\n", ans);
  }
}

### Prompt

Your challenge is to write 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 a, b, p, x;
long long Pow(long long x, long long y, long long mod) {
  if (!y) return 1;
  if (y == 1) return x % mod;
  long long t = Pow(x, y >> 1, mod), ans = t * t % mod;
  if (y & 1) ans = ans * x % mod;
  return ans;
}
int main() {
  while (scanf("%lld %lld %lld %lld", &a, &b, &p, &x) != EOF) {
    long long P = p * (p - 1), ans = 0;
    for (long long i = 1; i < p; i++) {
      long long r = b * Pow(Pow(a, i, p), p - 2, p) % p, c;
      if (i >= r)
        c = (i + (i - r) * (p - 1)) % P;
      else
        c = (i + (i + p - r) * (p - 1)) % P;
      if (!c) c = P;
      ans += ((x / P) + ((x % P >= c) ? 1 : 0));
    }
    printf("%lld\n", ans);
  }
}

```