#include <bits/stdc++.h>
using namespace std;
long long a, b, p, x;
long long modulo[2000000];
long long pwr(long long a, long long b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0) {
    long long temp = pwr(a, b / 2);
    return (temp * temp) % p;
  } else
    return (pwr(a, b - 1) * a) % p;
}
int main() {
  scanf("%lld %lld %lld %lld", &a, &b, &p, &x);
  a %= p;
  b %= p;
  long long temp = a;
  modulo[0] = 1;
  for (int i = 1; i <= p - 1; i++) {
    modulo[i] = temp % p;
    temp *= a;
    temp %= p;
  }
  for (int i = 0; i <= p - 1; i++) {
  }
  long long res = 0;
  for (long long i = 0; i < p - 1; i++) {
    long long X = b * pwr(modulo[i], p - 2);
    X %= p;
    long long ImX = i - X;
    ImX = ((ImX % p) + p) % p;
    long long A = p * p - p;
    long long B = ImX * (p - 1);
    long long temp = x - B - i;
    if (temp >= 0) {
      res += temp / A;
      res++;
    }
  }
  printf("%lld\n", res);
  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;
long long modulo[2000000];
long long pwr(long long a, long long b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0) {
    long long temp = pwr(a, b / 2);
    return (temp * temp) % p;
  } else
    return (pwr(a, b - 1) * a) % p;
}
int main() {
  scanf("%lld %lld %lld %lld", &a, &b, &p, &x);
  a %= p;
  b %= p;
  long long temp = a;
  modulo[0] = 1;
  for (int i = 1; i <= p - 1; i++) {
    modulo[i] = temp % p;
    temp *= a;
    temp %= p;
  }
  for (int i = 0; i <= p - 1; i++) {
  }
  long long res = 0;
  for (long long i = 0; i < p - 1; i++) {
    long long X = b * pwr(modulo[i], p - 2);
    X %= p;
    long long ImX = i - X;
    ImX = ((ImX % p) + p) % p;
    long long A = p * p - p;
    long long B = ImX * (p - 1);
    long long temp = x - B - i;
    if (temp >= 0) {
      res += temp / A;
      res++;
    }
  }
  printf("%lld\n", res);
  return 0;
}

```