#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long t = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') t = t * 10 + c - 48, c = getchar();
  return t * f;
}
vector<long long> vec[1000005];
long long a, b, p, x, ans, mo;
inline long long ksm(long long a, long long b) {
  long long t = 1, y = a;
  for (; b; b /= 2) {
    if (b & 1) t = 1ll * t * y % mo;
    y = 1ll * y * y % mo;
  }
  return t;
}
inline long long ni(long long a) { return ksm(a, mo - 2); }
inline long long get(long long b, long long z) {
  long long tt = b / (p - 1),
            ttt = upper_bound(vec[z].begin(), vec[z].end(), b % (p - 1)) -
                  vec[z].begin();
  return tt * vec[z].size() + ttt;
}
int main() {
  a = read();
  b = read();
  p = read();
  x = read();
  mo = p;
  long long t = a, ttp = 1;
  for (long long i = 0; i < p - 1; i++) {
    vec[ttp].push_back(i);
    ttp = ttp * t % p;
  }
  x++;
  for (long long i = 1; i < p; i++) {
    ans += get((x - i) / p + ((x - i) % p > 0) - 1,
               b * ni(i * ksm(a, i) % mo) % mo);
  }
  cout << ans << endl;
}

### Prompt

In CPP, your task is to solve 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;
inline long long read() {
  long long t = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') t = t * 10 + c - 48, c = getchar();
  return t * f;
}
vector<long long> vec[1000005];
long long a, b, p, x, ans, mo;
inline long long ksm(long long a, long long b) {
  long long t = 1, y = a;
  for (; b; b /= 2) {
    if (b & 1) t = 1ll * t * y % mo;
    y = 1ll * y * y % mo;
  }
  return t;
}
inline long long ni(long long a) { return ksm(a, mo - 2); }
inline long long get(long long b, long long z) {
  long long tt = b / (p - 1),
            ttt = upper_bound(vec[z].begin(), vec[z].end(), b % (p - 1)) -
                  vec[z].begin();
  return tt * vec[z].size() + ttt;
}
int main() {
  a = read();
  b = read();
  p = read();
  x = read();
  mo = p;
  long long t = a, ttp = 1;
  for (long long i = 0; i < p - 1; i++) {
    vec[ttp].push_back(i);
    ttp = ttp * t % p;
  }
  x++;
  for (long long i = 1; i < p; i++) {
    ans += get((x - i) / p + ((x - i) % p > 0) - 1,
               b * ni(i * ksm(a, i) % mo) % mo);
  }
  cout << ans << endl;
}

```