#include <bits/stdc++.h>
using namespace std;
const int MAXN = 300013, INF = 1 << 30;
long long a, b, p, x;
long long qmi(long long a, long long k, long long mod) {
  if (!k) return 1;
  if (k == 1) return a;
  long long t = qmi(a, k >> 1, mod);
  if (k & 1) return t * t % mod * a % mod;
  return t * t % mod;
}
void extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  extend_gcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - a / b * y;
}
int main() {
  cin >> a >> b >> p >> x;
  long long t, cx, cy, m, ans = 0;
  for (int i = 0; i < p - 1; i++) {
    t = qmi(a, i, p);
    extend_gcd(t, p, cx, cy);
    cx = ((cx + p) % p) * b % p;
    m = i * p - cx * p + cx;
    while (m <= 0) m += p * (p - 1);
    if (m <= x) ans += (x - m) / (p * (p - 1)) + 1;
  }
  cout << ans << endl;
  return 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;
const int MAXN = 300013, INF = 1 << 30;
long long a, b, p, x;
long long qmi(long long a, long long k, long long mod) {
  if (!k) return 1;
  if (k == 1) return a;
  long long t = qmi(a, k >> 1, mod);
  if (k & 1) return t * t % mod * a % mod;
  return t * t % mod;
}
void extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (!b) {
    x = 1;
    y = 0;
    return;
  }
  extend_gcd(b, a % b, x, y);
  long long t = x;
  x = y;
  y = t - a / b * y;
}
int main() {
  cin >> a >> b >> p >> x;
  long long t, cx, cy, m, ans = 0;
  for (int i = 0; i < p - 1; i++) {
    t = qmi(a, i, p);
    extend_gcd(t, p, cx, cy);
    cx = ((cx + p) % p) * b % p;
    m = i * p - cx * p + cx;
    while (m <= 0) m += p * (p - 1);
    if (m <= x) ans += (x - m) / (p * (p - 1)) + 1;
  }
  cout << ans << endl;
  return 0;
}

```