#include <bits/stdc++.h>
using namespace std;
long long pow1(long long b, long long expo, long long md) {
  if (!expo) return 1;
  long long tmp = pow1(b, expo / 2, md);
  tmp = (tmp * tmp) % md;
  if (expo % 2) tmp = (tmp * b) % md;
  return tmp;
}
long long w[1000005] = {};
long long apow[1000005] = {};
long long inv[1000005] = {};
long long v[1000005] = {};
long long get(long long a, long long m1, long long b, long long m2,
              long long x) {
  long long ret = 0;
  long long first, k;
  k = (a - b) * inv[m2];
  k %= m1;
  k += m1;
  k %= m1;
  first = k * m2 + b;
  first %= m1 * m2;
  x -= first;
  if (x >= 0) {
    ret += 1 + x / (m1 * m2);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  long long md, a, b, x, per;
  long long ans = 0;
  for (long long f = (long long)0; f <= (long long)1000004; ++f) v[f] = -1;
  cin >> a >> b >> md >> x;
  apow[0] = 1;
  v[1] = 0;
  per = 1;
  if (a != 1) {
    apow[1] = a;
    v[a] = 1;
    for (long long f = (long long)2; f <= (long long)md; ++f) {
      apow[f] = a * apow[f - 1];
      apow[f] %= md;
      if (apow[f] == 1) {
        per = f;
        break;
      }
      v[apow[f]] = f;
    }
  }
  for (long long f = (long long)1; f <= (long long)md - 1; ++f) {
    inv[f] = pow1(f, md - 2, md);
  }
  for (long long f = (long long)1; f <= (long long)md - 1; ++f) {
    w[f] = (inv[f] * b) % md;
  }
  for (long long f = (long long)1; f <= (long long)md - 1; ++f) {
    if (v[w[f]] != -1) {
      ans += get(f, md, v[w[f]], per, x);
    }
  }
  cout << ans;
  return 0;
}

### 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 pow1(long long b, long long expo, long long md) {
  if (!expo) return 1;
  long long tmp = pow1(b, expo / 2, md);
  tmp = (tmp * tmp) % md;
  if (expo % 2) tmp = (tmp * b) % md;
  return tmp;
}
long long w[1000005] = {};
long long apow[1000005] = {};
long long inv[1000005] = {};
long long v[1000005] = {};
long long get(long long a, long long m1, long long b, long long m2,
              long long x) {
  long long ret = 0;
  long long first, k;
  k = (a - b) * inv[m2];
  k %= m1;
  k += m1;
  k %= m1;
  first = k * m2 + b;
  first %= m1 * m2;
  x -= first;
  if (x >= 0) {
    ret += 1 + x / (m1 * m2);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(15);
  long long md, a, b, x, per;
  long long ans = 0;
  for (long long f = (long long)0; f <= (long long)1000004; ++f) v[f] = -1;
  cin >> a >> b >> md >> x;
  apow[0] = 1;
  v[1] = 0;
  per = 1;
  if (a != 1) {
    apow[1] = a;
    v[a] = 1;
    for (long long f = (long long)2; f <= (long long)md; ++f) {
      apow[f] = a * apow[f - 1];
      apow[f] %= md;
      if (apow[f] == 1) {
        per = f;
        break;
      }
      v[apow[f]] = f;
    }
  }
  for (long long f = (long long)1; f <= (long long)md - 1; ++f) {
    inv[f] = pow1(f, md - 2, md);
  }
  for (long long f = (long long)1; f <= (long long)md - 1; ++f) {
    w[f] = (inv[f] * b) % md;
  }
  for (long long f = (long long)1; f <= (long long)md - 1; ++f) {
    if (v[w[f]] != -1) {
      ans += get(f, md, v[w[f]], per, x);
    }
  }
  cout << ans;
  return 0;
}

```