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;
inline long long read() {
  register long long x = 0, f = 1;
  register char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
long long p, a, b, x, ans, inva, tmp, mini, maxi;
int Pow(long long a, long long b) {
  long long ans = 1;
  for (; b; b >>= 1, a = a * a % p)
    if (b & 1) ans = ans * a % p;
  return ans;
}
long long i, j;
int main() {
  a = read(), b = read(), p = read(), x = read();
  inva = Pow(a, p - 2);
  for (j = 0; j < p - 1; j++) {
    long long right = ((b * Pow(inva, j) - j) % p + p) % p;
    mini = (p - 1) * right % p;
    maxi = (x - j) / (p - 1);
    while (mini * (p - 1) + j > x) mini -= p;
    while (maxi * (p - 1) + j > x) maxi -= p;
    if (mini * (p - 1) + j <= 0) continue;
    ans += (maxi - mini + p) / p;
  }
  printf("%I64d\n", ans);
  return 0;
}