#include <bits/stdc++.h>
using namespace std;
const int N = 34;
long long a, b, p, x;
long long me(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long mi(long long n, long long mod) { return me(n, mod - 2, mod); }
int main() {
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &p);
  scanf("%lld", &x);
  long long ans = 0;
  for (long long i = 0; i < p - 1; i++) {
    long long here = me(a, i, p);
    here = mi(here, p);
    here = (here * b) % p;
    long long oh = here;
    long long rem = here - i;
    rem = (p - 1 - rem) % (p - 1);
    rem = (p * rem + oh);
    long long q = p * (p - 1);
    if (x >= rem) ans += 1 + (x - rem) / q;
  }
  printf("%lld", ans);
  return 0;
}

### Prompt

Please formulate 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;
const int N = 34;
long long a, b, p, x;
long long me(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long mi(long long n, long long mod) { return me(n, mod - 2, mod); }
int main() {
  scanf("%lld", &a);
  scanf("%lld", &b);
  scanf("%lld", &p);
  scanf("%lld", &x);
  long long ans = 0;
  for (long long i = 0; i < p - 1; i++) {
    long long here = me(a, i, p);
    here = mi(here, p);
    here = (here * b) % p;
    long long oh = here;
    long long rem = here - i;
    rem = (p - 1 - rem) % (p - 1);
    rem = (p * rem + oh);
    long long q = p * (p - 1);
    if (x >= rem) ans += 1 + (x - rem) / q;
  }
  printf("%lld", ans);
  return 0;
}

```