#include <bits/stdc++.h>
using namespace std;
long long gcd(long long n1, long long n2) {
  if (n2 == 0) return n1;
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long a, b, p, X;
long long powi[2000005];
long long powmod(long long base, long long exponent, long long MODi) {
  if (exponent < 0) exponent += MODi - 1;
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % MODi;
    base = (base * base) % MODi;
    exponent /= 2;
  }
  return ans;
}
int main() {
  cin >> a >> b >> p >> X;
  long long prod = p * (p - 1LL);
  powi[0] = 1;
  for (int i = 1; i <= p; i++) {
    powi[i] = a * powi[i - 1];
    powi[i] %= p;
  }
  long long ans = 0;
  for (int i = 0; i < p - 1; i++) {
    long long temp = b;
    temp *= powmod(powi[i], -1LL, p);
    temp %= p;
    long long x = powmod(p - 1, -1, p);
    x *= (p - 1LL);
    x %= prod;
    x *= temp;
    x %= prod;
    long long y = powmod(p, -1, p - 1LL);
    y *= (p);
    y %= prod;
    y *= i;
    y %= prod;
    x += y;
    x %= prod;
    if (x > 0 && X % prod >= x) ans++;
    ans += X / prod;
  }
  cout << ans << '\n';
}

### Prompt

Your challenge is to write 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 gcd(long long n1, long long n2) {
  if (n2 == 0) return n1;
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long a, b, p, X;
long long powi[2000005];
long long powmod(long long base, long long exponent, long long MODi) {
  if (exponent < 0) exponent += MODi - 1;
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % MODi;
    base = (base * base) % MODi;
    exponent /= 2;
  }
  return ans;
}
int main() {
  cin >> a >> b >> p >> X;
  long long prod = p * (p - 1LL);
  powi[0] = 1;
  for (int i = 1; i <= p; i++) {
    powi[i] = a * powi[i - 1];
    powi[i] %= p;
  }
  long long ans = 0;
  for (int i = 0; i < p - 1; i++) {
    long long temp = b;
    temp *= powmod(powi[i], -1LL, p);
    temp %= p;
    long long x = powmod(p - 1, -1, p);
    x *= (p - 1LL);
    x %= prod;
    x *= temp;
    x %= prod;
    long long y = powmod(p, -1, p - 1LL);
    y *= (p);
    y %= prod;
    y *= i;
    y %= prod;
    x += y;
    x %= prod;
    if (x > 0 && X % prod >= x) ans++;
    ans += X / prod;
  }
  cout << ans << '\n';
}

```