#include <bits/stdc++.h>
using namespace std;
long long mod, ans;
long long quickpow(long long a, long long b) {
  if (b < 0) return 0;
  long long ret = 1;
  a %= mod;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return ret;
}
long long inv(long long a) { return quickpow(a, mod - 2); }
int main() {
  long long a, b, p, x;
  while (~scanf("%lld %lld %lld %lld", &a, &b, &p, &x)) {
    mod = p;
    ans = 0;
    long long P = p * (p - 1);
    for (long long j = 0; j < p - 1; j++) {
      long long temp = b * inv(quickpow(a, j)) % p;
      long long i = (j - temp + p) % p;
      long long n = (i * (p - 1) + j) % P;
      if (n == 0) n = P;
      ans += x / P;
      if (x % P >= n) {
        ans++;
      }
    }
    printf("%lld\n", ans);
  }
}

### Prompt

Your challenge is to write 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 mod, ans;
long long quickpow(long long a, long long b) {
  if (b < 0) return 0;
  long long ret = 1;
  a %= mod;
  while (b) {
    if (b & 1) ret = (ret * a) % mod;
    b >>= 1;
    a = (a * a) % mod;
  }
  return ret;
}
long long inv(long long a) { return quickpow(a, mod - 2); }
int main() {
  long long a, b, p, x;
  while (~scanf("%lld %lld %lld %lld", &a, &b, &p, &x)) {
    mod = p;
    ans = 0;
    long long P = p * (p - 1);
    for (long long j = 0; j < p - 1; j++) {
      long long temp = b * inv(quickpow(a, j)) % p;
      long long i = (j - temp + p) % p;
      long long n = (i * (p - 1) + j) % P;
      if (n == 0) n = P;
      ans += x / P;
      if (x % P >= n) {
        ans++;
      }
    }
    printf("%lld\n", ans);
  }
}

```