#include <bits/stdc++.h>
using namespace std;
const double PI = 3.14159265358979;
const double eps = 1e-2;
const long long _INF = 1e18;
const int INF = 1e9;
const int maxn = 1000010;
const int maxm = 10000000;
long long a, b, p, x, m, q, t, r, k, ans;
long long kpow(long long a, long long b, long long p) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return ret;
}
int main() {
  while (~scanf("%I64d %I64d %I64d %I64d", &a, &b, &p, &x)) {
    ans = 0;
    m = p * (p - 1);
    q = x / m;
    t = x % m;
    r = b;
    for (long long i = p - 1; i > 0; --i) {
      k = (i * p - r * kpow(kpow(a, i, p), p - 2, p) % p * (p - 1)) % m;
      k = (k + m) % m;
      if (t >= k) ++ans;
    }
    ans += (p - 1) * q;
    printf("%I64d\n", 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;
const double PI = 3.14159265358979;
const double eps = 1e-2;
const long long _INF = 1e18;
const int INF = 1e9;
const int maxn = 1000010;
const int maxm = 10000000;
long long a, b, p, x, m, q, t, r, k, ans;
long long kpow(long long a, long long b, long long p) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return ret;
}
int main() {
  while (~scanf("%I64d %I64d %I64d %I64d", &a, &b, &p, &x)) {
    ans = 0;
    m = p * (p - 1);
    q = x / m;
    t = x % m;
    r = b;
    for (long long i = p - 1; i > 0; --i) {
      k = (i * p - r * kpow(kpow(a, i, p), p - 2, p) % p * (p - 1)) % m;
      k = (k + m) % m;
      if (t >= k) ++ans;
    }
    ans += (p - 1) * q;
    printf("%I64d\n", ans);
  }
  return 0;
}

```