#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000007;
const long long SZ = 3e5 + 1;
long long p;
long long fpow(long long a, long long k) {
  if (k == 0) return 1;
  if (k % 2 == 1)
    return (fpow(a, k - 1) * (long long)a) % p;
  else
    return fpow(((long long)a * a) % p, k / 2);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  long long a, b, x, ans = 0;
  cin >> a >> b >> p >> x;
  for (long long i = 0; i < min(p - 1, x + 1); i++) {
    long long v = fpow(a, i);
    long long m = b * fpow(v, p - 2) % p;
    long long u = (i + p - m) % p;
    long long vl = ((x - i) / (p - 1) - u);
    if (vl < 0) continue;
    ans += vl / p + 1;
  }
  cout << ans;
  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 long long INF = 1000000007;
const long long SZ = 3e5 + 1;
long long p;
long long fpow(long long a, long long k) {
  if (k == 0) return 1;
  if (k % 2 == 1)
    return (fpow(a, k - 1) * (long long)a) % p;
  else
    return fpow(((long long)a * a) % p, k / 2);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  long long a, b, x, ans = 0;
  cin >> a >> b >> p >> x;
  for (long long i = 0; i < min(p - 1, x + 1); i++) {
    long long v = fpow(a, i);
    long long m = b * fpow(v, p - 2) % p;
    long long u = (i + p - m) % p;
    long long vl = ((x - i) / (p - 1) - u);
    if (vl < 0) continue;
    ans += vl / p + 1;
  }
  cout << ans;
  return 0;
}

```