#include <bits/stdc++.h>
using namespace std;
long long n, Ans;
int a, inva, b, p, tot;
int A, B;
int f[1000010];
inline int Pow(int x, int y) {
  int Ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % p)
    if (y & 1) Ans = 1ll * Ans * x % p;
  return Ans;
}
inline long long Calc(int x, int y) {
  if (y == -1) return 0;
  long long r = (n - x) / p;
  if (r < tot) return (y <= r);
  return (r - y) / tot + 1;
}
int main() {
  int tttt = 1;
  scanf("%d%d%d%I64d", &a, &b, &p, &n);
  inva = Pow(a, p - 2);
  memset(f, -1, sizeof(f));
  int t = 1;
  for (int i = 0;; i++) {
    if (f[t] == -1)
      f[t] = i;
    else {
      tot = i;
      break;
    }
    t = 1ll * t * inva % p;
  }
  t = Pow(b, p - 2);
  for (int i = 0; i < p && i <= n; i++, t = 1ll * t * a % p) {
    int tmp = 1ll * i * t % p;
    Ans += Calc(i, f[tmp]);
  }
  cout << Ans << endl;
  return 0;
}

### 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 n, Ans;
int a, inva, b, p, tot;
int A, B;
int f[1000010];
inline int Pow(int x, int y) {
  int Ans = 1;
  for (; y; y >>= 1, x = 1ll * x * x % p)
    if (y & 1) Ans = 1ll * Ans * x % p;
  return Ans;
}
inline long long Calc(int x, int y) {
  if (y == -1) return 0;
  long long r = (n - x) / p;
  if (r < tot) return (y <= r);
  return (r - y) / tot + 1;
}
int main() {
  int tttt = 1;
  scanf("%d%d%d%I64d", &a, &b, &p, &n);
  inva = Pow(a, p - 2);
  memset(f, -1, sizeof(f));
  int t = 1;
  for (int i = 0;; i++) {
    if (f[t] == -1)
      f[t] = i;
    else {
      tot = i;
      break;
    }
    t = 1ll * t * inva % p;
  }
  t = Pow(b, p - 2);
  for (int i = 0; i < p && i <= n; i++, t = 1ll * t * a % p) {
    int tmp = 1ll * i * t % p;
    Ans += Calc(i, f[tmp]);
  }
  cout << Ans << endl;
  return 0;
}

```