#include <bits/stdc++.h>
using namespace std;
int i, j, k, cnt = 0;
long long qp(long long a, long long b, long long p) {
  long long r = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1ll) r = r * a % p;
  return r;
}
int main() {
  long long a, b, p, x, y, ans = 0;
  cin >> a >> b >> p >> x;
  long long nowa = a;
  for (i = 1; i < p; i++) {
    if (x - i < 0) {
      nowa = nowa * a % p;
      continue;
    }
    long long maxi = (x - i) / (p - 1);
    y = b * qp(nowa, p - 2, p) % p;
    long long l = (y - i) / p;
    if (p * l < y - i) l++;
    long long r = (y - i + maxi) / p;
    if (p * r > y - i + maxi) r--;
    ans += r - l + 1;
    nowa = a * nowa % p;
  }
  cout << ans << endl;
  return ~~(0 ^ 0 ^ 0);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 i, j, k, cnt = 0;
long long qp(long long a, long long b, long long p) {
  long long r = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1ll) r = r * a % p;
  return r;
}
int main() {
  long long a, b, p, x, y, ans = 0;
  cin >> a >> b >> p >> x;
  long long nowa = a;
  for (i = 1; i < p; i++) {
    if (x - i < 0) {
      nowa = nowa * a % p;
      continue;
    }
    long long maxi = (x - i) / (p - 1);
    y = b * qp(nowa, p - 2, p) % p;
    long long l = (y - i) / p;
    if (p * l < y - i) l++;
    long long r = (y - i + maxi) / p;
    if (p * r > y - i + maxi) r--;
    ans += r - l + 1;
    nowa = a * nowa % p;
  }
  cout << ans << endl;
  return ~~(0 ^ 0 ^ 0);
}

```