#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
long long a, b, p, x;
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int main() {
  cin >> a >> b >> p >> x;
  long long pp = 1, sum = 0;
  for (int i = 0; i < p - 1; i++) {
    long long x1, y, x2;
    long long g = gcdExtended(pp, p, &x1, &y);
    x1 %= p;
    x1 += p;
    x1 %= p;
    x1 *= b;
    x1 %= p;
    long long g1 = gcdExtended(p, p - 1, &x2, &y);
    long long t0 = -y * (i - x1);
    long long po = p * p - p;
    long long aa = (i + t0 * (p - 1)) % po;
    if (aa < 0) aa += po;
    sum += (x - aa) / po;
    if (aa) sum += aa <= x;
    pp *= a;
    pp %= p;
  }
  cout << sum;
}

### Prompt

Generate 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;
const int N = 2e5 + 7;
long long a, b, p, x;
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
int main() {
  cin >> a >> b >> p >> x;
  long long pp = 1, sum = 0;
  for (int i = 0; i < p - 1; i++) {
    long long x1, y, x2;
    long long g = gcdExtended(pp, p, &x1, &y);
    x1 %= p;
    x1 += p;
    x1 %= p;
    x1 *= b;
    x1 %= p;
    long long g1 = gcdExtended(p, p - 1, &x2, &y);
    long long t0 = -y * (i - x1);
    long long po = p * p - p;
    long long aa = (i + t0 * (p - 1)) % po;
    if (aa < 0) aa += po;
    sum += (x - aa) / po;
    if (aa) sum += aa <= x;
    pp *= a;
    pp %= p;
  }
  cout << sum;
}

```