#include <bits/stdc++.h>
using namespace std;
long long binpow(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  while (b) {
    if ((b & 1)) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  vector<long long> cnt(p, 0);
  long long aa = a;
  for (long long i = 1; i <= p; i++) {
    long long val = (aa * i) % p;
    cnt[val]++;
    aa *= a;
    aa %= p;
  }
  long long whole_cycle = x / p;
  long long partial_cycle = x % p;
  vector<long long> cycle(p - 1);
  cycle[0] = cnt[b];
  aa = a;
  for (long long i = 1; i < p - 1; i++) {
    long long who = (b * binpow(aa, p - 2, p)) % p;
    cycle[i] = cnt[who];
    cycle[i] += cycle[i - 1];
    aa *= a;
    aa %= p;
  }
  long long ans = cycle[p - 2] * (whole_cycle / (p - 1));
  if (whole_cycle % (p - 1) != 0) ans += cycle[(whole_cycle % (p - 1)) - 1];
  long long st = binpow(a, x - partial_cycle, p);
  for (long long i = x - partial_cycle + 1; i <= x; i++) {
    st *= a;
    st %= p;
    long long val = st * i;
    val %= p;
    if (val == b) ans++;
  }
  cout << ans;
}

### Prompt

Generate 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;
long long binpow(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  while (b) {
    if ((b & 1)) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
signed main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  long long a, b, p, x;
  cin >> a >> b >> p >> x;
  vector<long long> cnt(p, 0);
  long long aa = a;
  for (long long i = 1; i <= p; i++) {
    long long val = (aa * i) % p;
    cnt[val]++;
    aa *= a;
    aa %= p;
  }
  long long whole_cycle = x / p;
  long long partial_cycle = x % p;
  vector<long long> cycle(p - 1);
  cycle[0] = cnt[b];
  aa = a;
  for (long long i = 1; i < p - 1; i++) {
    long long who = (b * binpow(aa, p - 2, p)) % p;
    cycle[i] = cnt[who];
    cycle[i] += cycle[i - 1];
    aa *= a;
    aa %= p;
  }
  long long ans = cycle[p - 2] * (whole_cycle / (p - 1));
  if (whole_cycle % (p - 1) != 0) ans += cycle[(whole_cycle % (p - 1)) - 1];
  long long st = binpow(a, x - partial_cycle, p);
  for (long long i = x - partial_cycle + 1; i <= x; i++) {
    st *= a;
    st %= p;
    long long val = st * i;
    val %= p;
    if (val == b) ans++;
  }
  cout << ans;
}

```