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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
using namespace std;
string tostr(int n) {
  stringstream rr;
  rr << n;
  return rr.str();
}
const int mxn = 1e5 + 9;
const long double eps = 1e-9;
long long qpow(long long n, long long k, long long mod) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = (ans * n) % mod;
    n = (n * n) % mod;
    k >>= 1;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  long long i, j, k, n, m, a, b, x, val, p, t, mod, ans = 0;
  cin >> a >> b >> p >> x;
  mod = p * (p - 1);
  for (i = 0; i < p - 1; i++) {
    val = b * qpow(qpow(a, i, p), p - 2, p) % p;
    t = (mod + (i - val) * p % mod + val) % mod;
    if (x >= t) ans += 1 + (x - t) / mod;
  }
  cout << ans << "\n";
  return 0;
}