#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 7;
long long modexp(long long a, long long ex, long long md) {
  long long ans = 1;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * a) % md;
    ex >>= 1ll;
    a = (a * a) % md;
  }
  return ans;
}
vector<long long> mods;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  for (long long k = 0; k < p - 1; k++) {
    long long c = modexp(a, k, p);
    long long q = modexp(c, p - 2, p);
    q = (q * b) % p;
    long long z = (q * (p - 1) * (p - 1)) % (p * (p - 1));
    z = (z + k * p) % (p * (p - 1));
    mods.push_back(z);
  }
  long long s = p * (p - 1);
  sort(mods.begin(), mods.end());
  mods.resize(unique(mods.begin(), mods.end()) - mods.begin());
  long long ans = 0;
  for (long long y : mods) {
    if (x < y) continue;
    ans += (x - y) / s + 1;
  }
  cout << ans << endl;
}

### Prompt

Your challenge is to write 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 = 2e5 + 7;
long long modexp(long long a, long long ex, long long md) {
  long long ans = 1;
  while (ex > 0) {
    if (ex & 1ll) ans = (ans * a) % md;
    ex >>= 1ll;
    a = (a * a) % md;
  }
  return ans;
}
vector<long long> mods;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  for (long long k = 0; k < p - 1; k++) {
    long long c = modexp(a, k, p);
    long long q = modexp(c, p - 2, p);
    q = (q * b) % p;
    long long z = (q * (p - 1) * (p - 1)) % (p * (p - 1));
    z = (z + k * p) % (p * (p - 1));
    mods.push_back(z);
  }
  long long s = p * (p - 1);
  sort(mods.begin(), mods.end());
  mods.resize(unique(mods.begin(), mods.end()) - mods.begin());
  long long ans = 0;
  for (long long y : mods) {
    if (x < y) continue;
    ans += (x - y) / s + 1;
  }
  cout << ans << endl;
}

```