#include <bits/stdc++.h>
using namespace std;
long long qmul(long long a, long long b, long long P) {
  long long L = a * (b >> 25LL) % P * (1LL << 25) % P;
  long long R = a * (b & ((1LL << 25) - 1)) % P;
  return (L + R) % P;
}
long long qpow(long long a, long long b, long long p) {
  long long res = 1 % p;
  while (b) {
    if (b & 1) res = qmul(res, a, p);
    a = qmul(a, a, p);
    b >>= 1;
  }
  return res;
}
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  long long res = 0;
  a = qpow(a, p - 2, p);
  long long dw = p * (p - 1);
  for (int i = 0; i <= p - 2; ++i) {
    long long gamma = ((i - b) % p + p) % p;
    long long L = (1 - (p - 1) * gamma - i + dw - 1) / dw;
    long long R = floor((1.0 * (x - i) / (p - 1) - gamma) / p);
    b = a * b % p;
    res += R - L + 1;
  }
  printf("%lld", res);
  return 0;
}

### Prompt

Create a solution in cpp for 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 qmul(long long a, long long b, long long P) {
  long long L = a * (b >> 25LL) % P * (1LL << 25) % P;
  long long R = a * (b & ((1LL << 25) - 1)) % P;
  return (L + R) % P;
}
long long qpow(long long a, long long b, long long p) {
  long long res = 1 % p;
  while (b) {
    if (b & 1) res = qmul(res, a, p);
    a = qmul(a, a, p);
    b >>= 1;
  }
  return res;
}
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  long long res = 0;
  a = qpow(a, p - 2, p);
  long long dw = p * (p - 1);
  for (int i = 0; i <= p - 2; ++i) {
    long long gamma = ((i - b) % p + p) % p;
    long long L = (1 - (p - 1) * gamma - i + dw - 1) / dw;
    long long R = floor((1.0 * (x - i) / (p - 1) - gamma) / p);
    b = a * b % p;
    res += R - L + 1;
  }
  printf("%lld", res);
  return 0;
}

```