#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
long long mod, a, b, u;
long long fpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = 1ll * x * x % mod;
    n >>= 1;
  }
  return res;
}
int main() {
  scanf("%lld%lld%lld%lld", &a, &b, &mod, &u);
  long long inva = fpow(a, mod - 2), temp;
  temp = b * inva % mod;
  long long up = mod * (mod - 1);
  long long ans = 0;
  for (long long i = 1; i < mod; i++) {
    long long c;
    if (temp <= i)
      c = (i + (i - temp) * (mod - 1)) % up;
    else
      c = (i + (-temp + mod + i) * (mod - 1)) % up;
    if (!c) c = up;
    ans += (u / up + ((u % up >= c) ? 1 : 0));
    temp = temp * inva % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Generate 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;
const int INF = 0x3f3f3f3f;
long long mod, a, b, u;
long long fpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res = res * x % mod;
    x = 1ll * x * x % mod;
    n >>= 1;
  }
  return res;
}
int main() {
  scanf("%lld%lld%lld%lld", &a, &b, &mod, &u);
  long long inva = fpow(a, mod - 2), temp;
  temp = b * inva % mod;
  long long up = mod * (mod - 1);
  long long ans = 0;
  for (long long i = 1; i < mod; i++) {
    long long c;
    if (temp <= i)
      c = (i + (i - temp) * (mod - 1)) % up;
    else
      c = (i + (-temp + mod + i) * (mod - 1)) % up;
    if (!c) c = up;
    ans += (u / up + ((u % up >= c) ? 1 : 0));
    temp = temp * inva % mod;
  }
  printf("%lld\n", ans);
  return 0;
}

```