#include <bits/stdc++.h>
using namespace std;
const long long N = 3e6 + 5;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long t = 1, a, b, mod, x, num[2], rem[2];
long long inv(long long x, long long mod) {
  long long y = mod - 2;
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
long long findMinX() {
  long long prod = 1;
  for (int i = 0; i < 2; i++) prod = prod * num[i];
  long long result = 0;
  for (int i = 0; i < 2; i++) {
    long long pp = prod / num[i];
    result = result + rem[i] * inv(pp, num[i]) * pp;
  }
  return result % prod;
}
void solve() {
  cin >> a >> b >> mod >> x;
  long long p = mod * (mod - 1);
  long long q = x / p, r = x % p, res = 1, ans = 0;
  num[0] = mod, num[1] = mod - 1;
  for (long long i = 1; i < mod; i++) {
    res = (res * a) % mod;
    long long n = (b * inv(res, mod)) % mod;
    rem[0] = n, rem[1] = i % (mod - 1);
    long long X = findMinX();
    ans += q;
    if (X <= r) ans += 1;
  }
  cout << ans << endl;
}
int main() {
  while (t--) {
    solve();
  }
}

### Prompt

Please formulate 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 long long N = 3e6 + 5;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long t = 1, a, b, mod, x, num[2], rem[2];
long long inv(long long x, long long mod) {
  long long y = mod - 2;
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    x = (x * x) % mod;
    y /= 2;
  }
  return res;
}
long long findMinX() {
  long long prod = 1;
  for (int i = 0; i < 2; i++) prod = prod * num[i];
  long long result = 0;
  for (int i = 0; i < 2; i++) {
    long long pp = prod / num[i];
    result = result + rem[i] * inv(pp, num[i]) * pp;
  }
  return result % prod;
}
void solve() {
  cin >> a >> b >> mod >> x;
  long long p = mod * (mod - 1);
  long long q = x / p, r = x % p, res = 1, ans = 0;
  num[0] = mod, num[1] = mod - 1;
  for (long long i = 1; i < mod; i++) {
    res = (res * a) % mod;
    long long n = (b * inv(res, mod)) % mod;
    rem[0] = n, rem[1] = i % (mod - 1);
    long long X = findMinX();
    ans += q;
    if (X <= r) ans += 1;
  }
  cout << ans << endl;
}
int main() {
  while (t--) {
    solve();
  }
}

```