#include <bits/stdc++.h>
using namespace std;
long long mod;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long g = gcd(b % a, a, x1, y1);
  y = x1;
  x = y1 - x1 * (b / a);
  return g;
}
bool get_solution(long long a, long long b, long long c, long long &x,
                  long long &y, long long &g) {
  g = gcd(abs(a), abs(b), x, y);
  if (c % g != 0) {
    return 0;
  }
  x *= (c / g);
  y *= (c / g);
  if (a < 0) x *= -1;
  if (b < 0) y *= -1;
  return 1;
}
long long sqr(long long x) { return x * x; }
long long get_pow(long long a, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    return sqr(get_pow(a, p / 2)) % mod;
  } else {
    return a * (get_pow(a, p - 1)) % mod;
  }
}
signed main() {
  long long a, b, p, v;
  long long ans = 0;
  cin >> a >> b >> p >> v;
  for (long long i = 0; i < p - 1; i++) {
    mod = p;
    long long j = get_pow(a, i), x, y, g;
    if (!get_solution(j, -p, 1, x, y, g)) {
      continue;
    }
    long long k = (b * x) % p;
    if (!get_solution(p, -p + 1, i - k, x, y, g)) continue;
    long long step = (p - 1) / g * p;
    long long n = x * p + k;
    if (n >= 1) {
      n -= step * ((n - 1) / step);
    }
    if (n < 1) n += step * ((1 - n) / step + ((1 - n) % step != 0));
    ans += max((long long)0, (v - n) / step + (n <= v));
  }
  cout << ans;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
long long gcd(long long a, long long b, long long &x, long long &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long g = gcd(b % a, a, x1, y1);
  y = x1;
  x = y1 - x1 * (b / a);
  return g;
}
bool get_solution(long long a, long long b, long long c, long long &x,
                  long long &y, long long &g) {
  g = gcd(abs(a), abs(b), x, y);
  if (c % g != 0) {
    return 0;
  }
  x *= (c / g);
  y *= (c / g);
  if (a < 0) x *= -1;
  if (b < 0) y *= -1;
  return 1;
}
long long sqr(long long x) { return x * x; }
long long get_pow(long long a, long long p) {
  if (p == 0) return 1;
  if (p % 2 == 0) {
    return sqr(get_pow(a, p / 2)) % mod;
  } else {
    return a * (get_pow(a, p - 1)) % mod;
  }
}
signed main() {
  long long a, b, p, v;
  long long ans = 0;
  cin >> a >> b >> p >> v;
  for (long long i = 0; i < p - 1; i++) {
    mod = p;
    long long j = get_pow(a, i), x, y, g;
    if (!get_solution(j, -p, 1, x, y, g)) {
      continue;
    }
    long long k = (b * x) % p;
    if (!get_solution(p, -p + 1, i - k, x, y, g)) continue;
    long long step = (p - 1) / g * p;
    long long n = x * p + k;
    if (n >= 1) {
      n -= step * ((n - 1) / step);
    }
    if (n < 1) n += step * ((1 - n) / step + ((1 - n) % step != 0));
    ans += max((long long)0, (v - n) / step + (n <= v));
  }
  cout << ans;
  return 0;
}

```