#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000006;
int a, b, p;
long long x;
int cnt[maxn];
int A[maxn], B[maxn], C[maxn];
int main() {
  cin >> a >> b >> p >> x;
  B[0] = 1;
  int inva = 1;
  for (int i = 1; i < p - 1; i++) {
    inva = 1LL * inva * a % p;
  }
  for (int i = 1; i <= p; i++) {
    A[i] = i % p;
    B[i] = (B[i - 1] * 1LL * a) % p;
    C[i] = A[i] * 1LL * B[i] % p;
  }
  long long C1 = x / p;
  long long tmp = C1 / (p - 1);
  C1 = C1 % (p - 1);
  int C2 = x % p;
  for (int i = 1; i <= p; i++) {
    cnt[C[i]]++;
  }
  int goal = b;
  long long ans1 = 0, ans2 = 0;
  for (int i = 0; i < p - 1; i++) {
    ans2 += cnt[goal];
    goal = 1LL * goal * inva % p;
  }
  for (int i = 0; i < (int)C1; i++) {
    ans1 += cnt[goal];
    goal = 1LL * goal * inva % p;
  }
  long long ans = ans2 * tmp + ans1;
  for (int i = 1; i <= C2; i++) {
    if (C[i] == goal) ans++;
  }
  cout << ans << endl;
  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>
using namespace std;
const int maxn = 1000006;
int a, b, p;
long long x;
int cnt[maxn];
int A[maxn], B[maxn], C[maxn];
int main() {
  cin >> a >> b >> p >> x;
  B[0] = 1;
  int inva = 1;
  for (int i = 1; i < p - 1; i++) {
    inva = 1LL * inva * a % p;
  }
  for (int i = 1; i <= p; i++) {
    A[i] = i % p;
    B[i] = (B[i - 1] * 1LL * a) % p;
    C[i] = A[i] * 1LL * B[i] % p;
  }
  long long C1 = x / p;
  long long tmp = C1 / (p - 1);
  C1 = C1 % (p - 1);
  int C2 = x % p;
  for (int i = 1; i <= p; i++) {
    cnt[C[i]]++;
  }
  int goal = b;
  long long ans1 = 0, ans2 = 0;
  for (int i = 0; i < p - 1; i++) {
    ans2 += cnt[goal];
    goal = 1LL * goal * inva % p;
  }
  for (int i = 0; i < (int)C1; i++) {
    ans1 += cnt[goal];
    goal = 1LL * goal * inva % p;
  }
  long long ans = ans2 * tmp + ans1;
  for (int i = 1; i <= C2; i++) {
    if (C[i] == goal) ans++;
  }
  cout << ans << endl;
  return 0;
}

```