#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e6 + 10;
long long int a, b, c, d, e, f = 0, x, y, p, k, t, n, mp, ans = 0;
long long int po(long long int a, long long int b, long long int Q) {
  if (b == 0) return 1;
  long long int z = po(a, b / 2, Q);
  z *= z, z %= Q, z += Q, z %= Q;
  if (b % 2 == 1) z *= a;
  z %= Q, z += Q, z %= Q;
  return z;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> a >> b >> p >> x;
  mp = po(p, p - 3, p - 1);
  for (long long int d = 0; d < p - 1; d++) {
    if (d == 0)
      k = 1;
    else
      k *= a, k %= p;
    c = po(k, p - 2, p) * b, c %= p;
    e = d - c + p - 1, e %= (p - 1);
    e *= mp, e %= (p - 1);
    f = p * e + c;
    n = x - f;
    if (f > 0 && x >= f) ans++;
    if (n >= 0) n /= p * (p - 1), ans += n;
  }
  cout << ans;
}

### 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 long long int N = 1e6 + 10;
long long int a, b, c, d, e, f = 0, x, y, p, k, t, n, mp, ans = 0;
long long int po(long long int a, long long int b, long long int Q) {
  if (b == 0) return 1;
  long long int z = po(a, b / 2, Q);
  z *= z, z %= Q, z += Q, z %= Q;
  if (b % 2 == 1) z *= a;
  z %= Q, z += Q, z %= Q;
  return z;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> a >> b >> p >> x;
  mp = po(p, p - 3, p - 1);
  for (long long int d = 0; d < p - 1; d++) {
    if (d == 0)
      k = 1;
    else
      k *= a, k %= p;
    c = po(k, p - 2, p) * b, c %= p;
    e = d - c + p - 1, e %= (p - 1);
    e *= mp, e %= (p - 1);
    f = p * e + c;
    n = x - f;
    if (f > 0 && x >= f) ans++;
    if (n >= 0) n /= p * (p - 1), ans += n;
  }
  cout << ans;
}

```