#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e9 + 7;
long long power(long long x, long long y, long long m = MOD) {
  x %= m;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % m;
    x = x * x % m;
    y >>= 1;
  }
  return res;
}
long long modInv(long long x, long long m = MOD) { return power(x, m - 2); }
void run_case() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  ::MOD = p;
  long long res = 0, pow = 1;
  for (long long i = 0; i < p - 1; i++) {
    long long inv = power(pow, MOD - 2);
    long long sub = i - b * inv;
    sub = (sub % MOD + MOD) % MOD;
    res += ((x - i) / (MOD - 1) - sub) / MOD;
    if (i + sub * (MOD - 1) <= x) {
      res++;
    }
    pow = pow * a % MOD;
  }
  cout << res << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long T = 1;
  for (long long t = 1; t <= T; t++) {
    run_case();
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long MOD = 1e9 + 7;
long long power(long long x, long long y, long long m = MOD) {
  x %= m;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % m;
    x = x * x % m;
    y >>= 1;
  }
  return res;
}
long long modInv(long long x, long long m = MOD) { return power(x, m - 2); }
void run_case() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  ::MOD = p;
  long long res = 0, pow = 1;
  for (long long i = 0; i < p - 1; i++) {
    long long inv = power(pow, MOD - 2);
    long long sub = i - b * inv;
    sub = (sub % MOD + MOD) % MOD;
    res += ((x - i) / (MOD - 1) - sub) / MOD;
    if (i + sub * (MOD - 1) <= x) {
      res++;
    }
    pow = pow * a % MOD;
  }
  cout << res << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long T = 1;
  for (long long t = 1; t <= T; t++) {
    run_case();
  }
  return 0;
}

```