#include <bits/stdc++.h>
using namespace std;
long long po(long long a, long long x, long long mod) {
  if (x == 0) return 1LL;
  long long temp = po(a, x / 2, mod);
  temp = temp * temp % mod;
  if (x & 1) return a * temp % mod;
  return temp;
}
long long invmod(long long a, long long mod) { return po(a, mod - 2, mod); }
long long invpo(long long a, long long x, long long mod) {
  long long t = po(a, x, mod);
  return invmod(t, mod);
}
map<long long, long long> mark;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long ans = 0;
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  for (long long i = 0; i < p; i++) {
    long long aa = i;
    long long ab = b * invpo(a, i, p) % p;
    long long ap = p - 1;
    long long aq = p;
    long long ap1 = invmod(ap, aq);
    long long aq1 = invmod(aq, ap);
    long long N = p * (p - 1);
    long long an = (aa * aq * aq1 + ab * ap * ap1) % N;
    if (mark[an]) continue;
    if (an > x) continue;
    mark[an] = 1;
    ans += (x - an) / N + 1;
  }
  cout << ans << (char)0x0A;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 po(long long a, long long x, long long mod) {
  if (x == 0) return 1LL;
  long long temp = po(a, x / 2, mod);
  temp = temp * temp % mod;
  if (x & 1) return a * temp % mod;
  return temp;
}
long long invmod(long long a, long long mod) { return po(a, mod - 2, mod); }
long long invpo(long long a, long long x, long long mod) {
  long long t = po(a, x, mod);
  return invmod(t, mod);
}
map<long long, long long> mark;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long ans = 0;
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  for (long long i = 0; i < p; i++) {
    long long aa = i;
    long long ab = b * invpo(a, i, p) % p;
    long long ap = p - 1;
    long long aq = p;
    long long ap1 = invmod(ap, aq);
    long long aq1 = invmod(aq, ap);
    long long N = p * (p - 1);
    long long an = (aa * aq * aq1 + ab * ap * ap1) % N;
    if (mark[an]) continue;
    if (an > x) continue;
    mark[an] = 1;
    ans += (x - an) / N + 1;
  }
  cout << ans << (char)0x0A;
  return 0;
}

```