#include <bits/stdc++.h>
using namespace std;
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modInverse(long long a, long long m) {
  long long g = gcd(a, m);
  if (g != 1)
    return -1;
  else
    return power(a, m - 2, m);
}
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  long long pow = 1;
  long long cnt(0);
  for (long long j = 1; j < p; ++j) {
    pow = (pow * a) % p;
    long long invPow = modInverse(pow, p);
    if (invPow == -1) continue;
    long long y = (b * invPow) % p;
    long long mn, d;
    mn = ((j - y + p) % p) * (p - 1) + j;
    if (mn > x) continue;
    cnt += (x - mn) / (p * (p - 1)) + 1;
  }
  cout << cnt;
}

### Prompt

Please formulate 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 power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long modInverse(long long a, long long m) {
  long long g = gcd(a, m);
  if (g != 1)
    return -1;
  else
    return power(a, m - 2, m);
}
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  long long pow = 1;
  long long cnt(0);
  for (long long j = 1; j < p; ++j) {
    pow = (pow * a) % p;
    long long invPow = modInverse(pow, p);
    if (invPow == -1) continue;
    long long y = (b * invPow) % p;
    long long mn, d;
    mn = ((j - y + p) % p) * (p - 1) + j;
    if (mn > x) continue;
    cnt += (x - mn) / (p * (p - 1)) + 1;
  }
  cout << cnt;
}

```