#include <bits/stdc++.h>
using namespace std;
using ll = long long;
ll fastexp(ll x, ll y, ll p) {
  ll ans = 1;
  while (y > 0) {
    if (y & 1) ans = (1LL * ans * x) % p;
    y = y >> 1;
    x = (1LL * x * x) % p;
  }
  return ans % p;
}
ll invmod(ll x, ll p) { return fastexp(x, p - 2, p) % p; }
ll arrp[1000005];
int main() {
  int a, b, p;
  ll x;
  cin >> a >> b >> p >> x;
  arrp[0] = 1;
  int firstn;
  ll answ = 0;
  for (int i = 1; i < 1000005; i++) arrp[i] = (arrp[i - 1] * a) % p;
  for (int j = 0; j < p - 1; j++) {
    ll i = (1LL * b * invmod(arrp[j], p)) % p;
    i = (i - j) % p;
    i = (-1 * i + p) % p;
    answ += ((x - j) / (p - 1) - i) / p;
    if ((x - j) >= 1LL * i * (p - 1)) {
      answ++;
    }
  }
  cout << answ << '\n';
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
using ll = long long;
ll fastexp(ll x, ll y, ll p) {
  ll ans = 1;
  while (y > 0) {
    if (y & 1) ans = (1LL * ans * x) % p;
    y = y >> 1;
    x = (1LL * x * x) % p;
  }
  return ans % p;
}
ll invmod(ll x, ll p) { return fastexp(x, p - 2, p) % p; }
ll arrp[1000005];
int main() {
  int a, b, p;
  ll x;
  cin >> a >> b >> p >> x;
  arrp[0] = 1;
  int firstn;
  ll answ = 0;
  for (int i = 1; i < 1000005; i++) arrp[i] = (arrp[i - 1] * a) % p;
  for (int j = 0; j < p - 1; j++) {
    ll i = (1LL * b * invmod(arrp[j], p)) % p;
    i = (i - j) % p;
    i = (-1 * i + p) % p;
    answ += ((x - j) / (p - 1) - i) / p;
    if ((x - j) >= 1LL * i * (p - 1)) {
      answ++;
    }
  }
  cout << answ << '\n';
}

```