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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long bin_pow(long long x, long long e, long long MOD) {
  long long res = 1;
  while (e) {
    if (e & 1) res = res * x % MOD;
    e >>= 1;
    x = x * x % MOD;
  }
  return res;
}
long long p, a, b, x, ans;
vector<long long> m[1123456];
long long xd[1123456];
long long xy[1123456];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> a >> b >> p >> x;
  for (int i = 1; i < p; i++) {
    xd[i] = bin_pow(i, p - 2, p);
    xy[i] = bin_pow(a, i, p);
  }
  for (int i = 0; i < p - 1; i++) {
    m[bin_pow(a, i, p)].push_back(i);
  }
  for (int i = 1; i < p; i++) {
    vector<long long> g = m[((b * xd[i]) % p) * xd[xy[i]] % p];
    for (auto k : g) {
      long long y = x - i;
      if (y < 0) continue;
      y /= p;
      long long t = y / (p - 1);
      if (t * (p - 1) + k <= y)
        ans += t + 1;
      else
        ans += t;
    }
  }
  cout << ans;
  return 0;
}