#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b, long long m) {
  long long x = 1 % m;
  a %= m;
  while (b) {
    if ((1 & b)) x = x * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return x;
}
vector<long long> sol[1000005];
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  long long aa = 1;
  for (int q = 0; q < p - 1; q++) {
    sol[aa].push_back(q);
    aa = aa * a % p;
  }
  long long ans = 0;
  for (long long k = 1; k < p; k++) {
    if (k > x) break;
    long long d = b * power(k * power(a, k, p), p - 2, p) % p;
    if (sol[d].empty() == true) continue;
    long long lim = (x - k) / p;
    ans += ((int)sol[d].size()) * 1LL * (lim / (p - 1));
    long long upto = lim % (p - 1);
    ans += upper_bound(sol[d].begin(), sol[d].end(), upto) - sol[d].begin();
  }
  printf("%lld\n", ans);
}

### Prompt

Create a solution in CPP for 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 power(long long a, long long b, long long m) {
  long long x = 1 % m;
  a %= m;
  while (b) {
    if ((1 & b)) x = x * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return x;
}
vector<long long> sol[1000005];
int main() {
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  long long aa = 1;
  for (int q = 0; q < p - 1; q++) {
    sol[aa].push_back(q);
    aa = aa * a % p;
  }
  long long ans = 0;
  for (long long k = 1; k < p; k++) {
    if (k > x) break;
    long long d = b * power(k * power(a, k, p), p - 2, p) % p;
    if (sol[d].empty() == true) continue;
    long long lim = (x - k) / p;
    ans += ((int)sol[d].size()) * 1LL * (lim / (p - 1));
    long long upto = lim % (p - 1);
    ans += upper_bound(sol[d].begin(), sol[d].end(), upto) - sol[d].begin();
  }
  printf("%lld\n", ans);
}

```