#include <bits/stdc++.h>
using namespace std;
long long a, b, p;
long long x;
long long powMod(long long a, long long j, long long p) {
  long long ans = 1;
  while (j) {
    if (j & 1) {
      (ans *= a) %= p;
    }
    (a *= a) %= p;
    j /= 2;
  }
  return ans;
}
long long inv(long long y, long long p) { return powMod(y, p - 2, p); }
int main() {
  scanf("%lld %lld %lld %lld", &a, &b, &p, &x);
  long long ans = 0;
  if (a % p == 0)
    ans = x;
  else
    for (int j = 0; j < p - 1; j++) {
      long long curr = (long long)b * inv(powMod(a, j, p), p) % p;
      long long first = (long long)(p - 1) * ((j - curr + p) % p) + j;
      if (first > x) continue;
      ans += (x - first) / ((long long)p * (p - 1)) + 1;
    }
  printf("%lld \n", ans);
}

### 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;
long long a, b, p;
long long x;
long long powMod(long long a, long long j, long long p) {
  long long ans = 1;
  while (j) {
    if (j & 1) {
      (ans *= a) %= p;
    }
    (a *= a) %= p;
    j /= 2;
  }
  return ans;
}
long long inv(long long y, long long p) { return powMod(y, p - 2, p); }
int main() {
  scanf("%lld %lld %lld %lld", &a, &b, &p, &x);
  long long ans = 0;
  if (a % p == 0)
    ans = x;
  else
    for (int j = 0; j < p - 1; j++) {
      long long curr = (long long)b * inv(powMod(a, j, p), p) % p;
      long long first = (long long)(p - 1) * ((j - curr + p) % p) + j;
      if (first > x) continue;
      ans += (x - first) / ((long long)p * (p - 1)) + 1;
    }
  printf("%lld \n", ans);
}

```