#include <bits/stdc++.h>
using namespace std;
const int N = 10e6 + 5;
int inversa[N];
void CalcInv(int p) {
  inversa[1] = 1;
  for (int i = 2; i < p; i++) {
    inversa[i] = (p - p / i) * 1ll * inversa[p % i] % p;
  }
}
long long expbi(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  CalcInv(p);
  long long res = 0;
  for (int i = 1; i < p; i++) {
    long long y, ymax;
    y = ((b * inversa[expbi(a, i, p)] - i + p) % p * inversa[p - 1]) % p;
    ymax = 1ll * (x - i) / (p - 1);
    if (i + y * (p - 1) > x) continue;
    res += 1ll * (ymax - y) / p + 1;
  }
  cout << res << "\n";
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 10e6 + 5;
int inversa[N];
void CalcInv(int p) {
  inversa[1] = 1;
  for (int i = 2; i < p; i++) {
    inversa[i] = (p - p / i) * 1ll * inversa[p % i] % p;
  }
}
long long expbi(long long a, long long b, long long p) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  CalcInv(p);
  long long res = 0;
  for (int i = 1; i < p; i++) {
    long long y, ymax;
    y = ((b * inversa[expbi(a, i, p)] - i + p) % p * inversa[p - 1]) % p;
    ymax = 1ll * (x - i) / (p - 1);
    if (i + y * (p - 1) > x) continue;
    res += 1ll * (ymax - y) / p + 1;
  }
  cout << res << "\n";
  return 0;
}

```