#include <bits/stdc++.h>
using namespace std;
long long pw(long long a, long long b, long long p) {
  long long res = 1;
  a %= p;
  b %= p - 1;
  while (b) {
    if (b & 1) res = (1LL * a * res) % p;
    a = (1LL * a * a) % p;
    b >>= 1;
  }
  return res;
}
long long a, b, x, p, ans;
vector<pair<long long, long long> > pir;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> a >> b >> p >> x;
  for (long long i = 0; i < p - 1; i++) {
    long long t = pw(a, i, p);
    t = pw(t, p - 2, p);
    t = (1LL * b * t) % p;
    pir.push_back({i, t});
  }
  long long lcm = p * (p - 1);
  for (auto i : pir) {
    long long z = i.first - i.second;
    long long w = z * p + i.second;
    w %= lcm;
    if (w < 0) w += lcm;
    if (w > x) continue;
    ans += (x - w) / lcm + 1;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 pw(long long a, long long b, long long p) {
  long long res = 1;
  a %= p;
  b %= p - 1;
  while (b) {
    if (b & 1) res = (1LL * a * res) % p;
    a = (1LL * a * a) % p;
    b >>= 1;
  }
  return res;
}
long long a, b, x, p, ans;
vector<pair<long long, long long> > pir;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> a >> b >> p >> x;
  for (long long i = 0; i < p - 1; i++) {
    long long t = pw(a, i, p);
    t = pw(t, p - 2, p);
    t = (1LL * b * t) % p;
    pir.push_back({i, t});
  }
  long long lcm = p * (p - 1);
  for (auto i : pir) {
    long long z = i.first - i.second;
    long long w = z * p + i.second;
    w %= lcm;
    if (w < 0) w += lcm;
    if (w > x) continue;
    ans += (x - w) / lcm + 1;
  }
  cout << ans << endl;
  return 0;
}

```