#include <bits/stdc++.h>
using namespace std;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << '\t';
  err(++it, args...);
}
const int lim = 2000 + 10;
long long int pow(long long int base, long long int exp, int m) {
  if (exp == 0)
    return 1;
  else if (exp == 1)
    return base;
  long long int root = pow(base, exp >> 1, m);
  if ((exp & 1) == 0)
    return (root * root) % m;
  else
    return (((root * root) % m) * base) % m;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, p, x, n, t, i, j, h, y, X, ans = 0;
  cin >> a >> b >> p >> x;
  for (j = 1; j < p; j++) {
    y = pow(a, j * (p - 2), p);
    y = (y * b) % p;
    X = j - y;
    long long int lb = X < 0 ? 1 : 0;
    h = (x - j - X * (p - 1)) / (p - 1);
    h /= p;
    if ((p - 1) * (h * p + X) + j > x) h--;
    if (h - lb + 1 > 0) ans += (h - lb + 1);
  }
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) v.emplace_back(x);
  return move(v);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << '\t';
  err(++it, args...);
}
const int lim = 2000 + 10;
long long int pow(long long int base, long long int exp, int m) {
  if (exp == 0)
    return 1;
  else if (exp == 1)
    return base;
  long long int root = pow(base, exp >> 1, m);
  if ((exp & 1) == 0)
    return (root * root) % m;
  else
    return (((root * root) % m) * base) % m;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int a, b, p, x, n, t, i, j, h, y, X, ans = 0;
  cin >> a >> b >> p >> x;
  for (j = 1; j < p; j++) {
    y = pow(a, j * (p - 2), p);
    y = (y * b) % p;
    X = j - y;
    long long int lb = X < 0 ? 1 : 0;
    h = (x - j - X * (p - 1)) / (p - 1);
    h /= p;
    if ((p - 1) * (h * p + X) + j > x) h--;
    if (h - lb + 1 > 0) ans += (h - lb + 1);
  }
  cout << ans;
  return 0;
}

```