#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5;
long long a, b, p, X, ans, a_inv;
long long ksm(long long a, long long b) {
  long long t = 1;
  while (b) {
    if (b & 1) t = t * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return t;
}
long long up(long long x, long long y) {
  long long re = x / y;
  if (x > 0) re += x % y != 0;
  return re;
}
long long down(long long x, long long y) {
  long long re = x / y;
  if (x < 0) re -= x % y != 0;
  return re;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &a, &b, &p, &X);
  a_inv = ksm(a, p - 2);
  long long y = b, ans = 0;
  for (long long j = 1; j < p; j++) {
    y = y * a_inv % p;
    long long t = down(down(X - j, p - 1) + y - j, p) - up(y - j, p) + 1;
    t = max(t, 0ll);
    ans += t;
  }
  printf("%I64d", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 maxn = 5e5;
long long a, b, p, X, ans, a_inv;
long long ksm(long long a, long long b) {
  long long t = 1;
  while (b) {
    if (b & 1) t = t * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return t;
}
long long up(long long x, long long y) {
  long long re = x / y;
  if (x > 0) re += x % y != 0;
  return re;
}
long long down(long long x, long long y) {
  long long re = x / y;
  if (x < 0) re -= x % y != 0;
  return re;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &a, &b, &p, &X);
  a_inv = ksm(a, p - 2);
  long long y = b, ans = 0;
  for (long long j = 1; j < p; j++) {
    y = y * a_inv % p;
    long long t = down(down(X - j, p - 1) + y - j, p) - up(y - j, p) + 1;
    t = max(t, 0ll);
    ans += t;
  }
  printf("%I64d", ans);
  return 0;
}

```