#include <bits/stdc++.h>
using namespace std;
const int SIZE = 1000;
long long a, b, p, x;
long long qpow(long long x, long long y, long long MOD) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    y = y >> 1;
  }
  return res;
}
int main() {
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  long long ans = 0;
  for (long long i = 1; i < p; i++) {
    long long tp = qpow(a, i, p);
    long long invt = qpow(tp, p - 2, p);
    long long invp = qpow(p - 1, p - 2, p);
    long long y = ((b * invt - i + p) % p * invp) % p;
    long long ty = (x - i) / (p - 1);
    if (i + y * (p - 1) > x) continue;
    long long anst = (ty - y) / p + 1;
    ans = ans + anst;
  }
  printf("%lld\n", ans);
  return 0;
}

### 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;
const int SIZE = 1000;
long long a, b, p, x;
long long qpow(long long x, long long y, long long MOD) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % MOD;
    x = (x * x) % MOD;
    y = y >> 1;
  }
  return res;
}
int main() {
  scanf("%lld%lld%lld%lld", &a, &b, &p, &x);
  long long ans = 0;
  for (long long i = 1; i < p; i++) {
    long long tp = qpow(a, i, p);
    long long invt = qpow(tp, p - 2, p);
    long long invp = qpow(p - 1, p - 2, p);
    long long y = ((b * invt - i + p) % p * invp) % p;
    long long ty = (x - i) / (p - 1);
    if (i + y * (p - 1) > x) continue;
    long long anst = (ty - y) / p + 1;
    ans = ans + anst;
  }
  printf("%lld\n", ans);
  return 0;
}

```