#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x, i, d, pos;
long long an, act, rat, k;
long long sol;
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long cmmmc(long long a, long long b) { return (a * b) / gcd(a, b); }
long long poww(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return ans;
}
int main() {
  cin >> a >> b >> p >> x;
  an = a;
  for (i = 1; i <= p - 1 && i <= x; i++) {
    act = (an * i) % p;
    rat = p - an;
    rat %= p;
    k = poww(rat, p - 2);
    k = (k * (p + b - act)) % p;
    k *= (p - 1);
    pos = i + k;
    if (pos <= x) {
      sol++;
      sol += (x - pos) / ((p - 1) * p);
    }
    an = (an * a) % p;
  }
  cout << sol;
  return 0;
}

### Prompt

Please create a solution in CPP 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 a, b, p, x, i, d, pos;
long long an, act, rat, k;
long long sol;
long long gcd(long long a, long long b) {
  if (a < b) swap(a, b);
  while (b > 0) {
    a %= b;
    swap(a, b);
  }
  return a;
}
long long cmmmc(long long a, long long b) { return (a * b) / gcd(a, b); }
long long poww(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return ans;
}
int main() {
  cin >> a >> b >> p >> x;
  an = a;
  for (i = 1; i <= p - 1 && i <= x; i++) {
    act = (an * i) % p;
    rat = p - an;
    rat %= p;
    k = poww(rat, p - 2);
    k = (k * (p + b - act)) % p;
    k *= (p - 1);
    pos = i + k;
    if (pos <= x) {
      sol++;
      sol += (x - pos) / ((p - 1) * p);
    }
    an = (an * a) % p;
  }
  cout << sol;
  return 0;
}

```