#include <bits/stdc++.h>
int u[1200000], v[1200000];
long long d[1200000];
int i, k, p, x, y;
long long n, s, t;
inline int power(int x, int y) {
  int s = 1;
  for (int i = 30; i >= 0; i--) {
    s = 1LL * s * s % p;
    if ((1 << i) & y) s = 1LL * s * x % p;
  }
  return s;
}
int main() {
  scanf("%d%d%d%I64d", &x, &y, &p, &n);
  u[0] = 1;
  for (i = 1; i < p; i++) u[i] = 1LL * u[i - 1] * x % p;
  for (i = 1; i < p; i++) v[i] = 1LL * y * power(i, p - 2) % p;
  t = n / p;
  for (i = 1; i < p; i++) d[u[i]] = d[u[i]] + t / (p - 1);
  for (k = 1; k <= t % (p - 1); k++) d[u[k]]++;
  for (i = 1; i <= p; i++) {
    s = s + d[v[i]];
    d[u[i]]--;
    d[u[k]]++;
    k++;
    if (k >= p) k = 1;
  }
  for (i = 1, k = t % (p - 1); i <= n % p; i++) {
    k++;
    if (k >= p) k = 1;
    if (u[k] == v[i]) s++;
  }
  printf("%I64d", s);
  return 0;
}

### Prompt

Your task is to create 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>
int u[1200000], v[1200000];
long long d[1200000];
int i, k, p, x, y;
long long n, s, t;
inline int power(int x, int y) {
  int s = 1;
  for (int i = 30; i >= 0; i--) {
    s = 1LL * s * s % p;
    if ((1 << i) & y) s = 1LL * s * x % p;
  }
  return s;
}
int main() {
  scanf("%d%d%d%I64d", &x, &y, &p, &n);
  u[0] = 1;
  for (i = 1; i < p; i++) u[i] = 1LL * u[i - 1] * x % p;
  for (i = 1; i < p; i++) v[i] = 1LL * y * power(i, p - 2) % p;
  t = n / p;
  for (i = 1; i < p; i++) d[u[i]] = d[u[i]] + t / (p - 1);
  for (k = 1; k <= t % (p - 1); k++) d[u[k]]++;
  for (i = 1; i <= p; i++) {
    s = s + d[v[i]];
    d[u[i]]--;
    d[u[k]]++;
    k++;
    if (k >= p) k = 1;
  }
  for (i = 1, k = t % (p - 1); i <= n % p; i++) {
    k++;
    if (k >= p) k = 1;
    if (u[k] == v[i]) s++;
  }
  printf("%I64d", s);
  return 0;
}

```