#include <bits/stdc++.h>
using namespace std;
long long extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    long long r = extend_gcd(b, a % b, y, x);
    y -= x * (a / b);
    return r;
  }
}
int line_mod_equation(long long a, long long b, long long n) {
  long long x, y;
  long long d = extend_gcd(a, n, x, y);
  if (b % d == 0) {
    x %= n;
    x += n;
    x %= n;
    return x * (b / d) % (n / d);
  }
  return -1;
}
int main() {
  long long a, b, p, x;
  scanf("%lld", &a), scanf("%lld", &b), scanf("%lld", &p), scanf("%lld", &x);
  long long pre = 1;
  long long ans = 0;
  for (int i = 1; i <= p - 1; i++) {
    pre = pre * a % p;
    long long res = line_mod_equation(pre, b, p);
    if (res == -1) continue;
    long long L = i;
    long long R = L - x / (p - 1) + 1;
    if (x / (p - 1) * (p - 1) + i <= x) R--;
    swap(L, R);
    L += p;
    R += p;
    res = (L - res) / p * p + res;
    res -= 5 * p;
    while (res < L) res += p;
    long long dl = res;
    long long rr = (R - res) / p * p + res;
    while (rr + p <= R) rr += p;
    if (rr >= dl && rr <= R && dl >= L) ans += (rr - dl) / p + 1;
  }
  printf("%lld\n", ans);
  ;
  return 0;
}

### Prompt

Please formulate 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 extend_gcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  } else {
    long long r = extend_gcd(b, a % b, y, x);
    y -= x * (a / b);
    return r;
  }
}
int line_mod_equation(long long a, long long b, long long n) {
  long long x, y;
  long long d = extend_gcd(a, n, x, y);
  if (b % d == 0) {
    x %= n;
    x += n;
    x %= n;
    return x * (b / d) % (n / d);
  }
  return -1;
}
int main() {
  long long a, b, p, x;
  scanf("%lld", &a), scanf("%lld", &b), scanf("%lld", &p), scanf("%lld", &x);
  long long pre = 1;
  long long ans = 0;
  for (int i = 1; i <= p - 1; i++) {
    pre = pre * a % p;
    long long res = line_mod_equation(pre, b, p);
    if (res == -1) continue;
    long long L = i;
    long long R = L - x / (p - 1) + 1;
    if (x / (p - 1) * (p - 1) + i <= x) R--;
    swap(L, R);
    L += p;
    R += p;
    res = (L - res) / p * p + res;
    res -= 5 * p;
    while (res < L) res += p;
    long long dl = res;
    long long rr = (R - res) / p * p + res;
    while (rr + p <= R) rr += p;
    if (rr >= dl && rr <= R && dl >= L) ans += (rr - dl) / p + 1;
  }
  printf("%lld\n", ans);
  ;
  return 0;
}

```