#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
long long exgcd(long long a, long long b, long long &x, long long &y) {
  long long d = a;
  if (b == 0)
    x = 1, y = 0;
  else
    d = exgcd(b, a % b, y, x), y -= x * (a / b);
  return d;
}
long long solve(long long a, long long b, long long c, long long &x,
                long long &y, long long &dx, long long &dy) {
  long long d = exgcd(a, b, x, y);
  if (c % d) return false;
  dx = b / d, dy = a / d;
  if (dx < 0) dx = -dx, dy = -dy;
  x = (((x * c / d) % (dx) + (dx)) % (dx));
  y = (c - a * x) / b;
  return true;
}
long long qpow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans % mod;
}
int main() {
  long long a, p, b, x, ans = 0LL;
  cin >> a >> b >> p >> x;
  for (int i = 0; i < p - 1; i++) {
    long long X, Y, dx, dy;
    long long ar = qpow(a, i, p);
    if (solve(ar, p, i * ar - b, X, Y, dx, dy)) {
      long long n = i + X * (p - 1);
      if (x >= n) ans += (x - n) / (dx * (p - 1)) + 1;
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 = 1e5 + 10;
long long exgcd(long long a, long long b, long long &x, long long &y) {
  long long d = a;
  if (b == 0)
    x = 1, y = 0;
  else
    d = exgcd(b, a % b, y, x), y -= x * (a / b);
  return d;
}
long long solve(long long a, long long b, long long c, long long &x,
                long long &y, long long &dx, long long &dy) {
  long long d = exgcd(a, b, x, y);
  if (c % d) return false;
  dx = b / d, dy = a / d;
  if (dx < 0) dx = -dx, dy = -dy;
  x = (((x * c / d) % (dx) + (dx)) % (dx));
  y = (c - a * x) / b;
  return true;
}
long long qpow(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return ans % mod;
}
int main() {
  long long a, p, b, x, ans = 0LL;
  cin >> a >> b >> p >> x;
  for (int i = 0; i < p - 1; i++) {
    long long X, Y, dx, dy;
    long long ar = qpow(a, i, p);
    if (solve(ar, p, i * ar - b, X, Y, dx, dy)) {
      long long n = i + X * (p - 1);
      if (x >= n) ans += (x - n) / (dx * (p - 1)) + 1;
    }
  }
  cout << ans << endl;
  return 0;
}

```