#include <bits/stdc++.h>
using namespace std;
long long x, y, z, sol;
int a, b, p;
int exp(int w, int q) {
  if (q == 0) return 1;
  if (q == 1) return w;
  int e = exp(1LL * w * w % p, q / 2);
  if (q % 2 == 0) return e;
  return (1LL * e * w % p);
}
int euclid(int a, int b, int *x, int *y) {
  if (b == 0) {
    *x = 1;
    *y = 0;
    return a;
  }
  int x0, y0;
  int d = euclid(b, a % b, &x0, &y0);
  *x = y0;
  *y = x0 - (a / b) * y0;
  return d;
}
int inversmodular(int a) {
  int x, y;
  int d = euclid(a, p, &x, &y);
  return ((x + p) % p);
}
int main() {
  int i, j;
  scanf("%d %d %d %I64d", &a, &b, &p, &x);
  for (j = 1; j < p; j++) {
    y = j - 1LL * b * inversmodular(exp(a, j)) % p;
    y = (y + p) % p;
    if (x - j < 0) continue;
    z = (x - j) / (p - 1);
    z -= y;
    if (z >= 0) {
      sol += z / p + 1;
    }
  }
  printf("%I64d", sol);
  return 0;
}

### Prompt

Create a solution in Cpp for 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 x, y, z, sol;
int a, b, p;
int exp(int w, int q) {
  if (q == 0) return 1;
  if (q == 1) return w;
  int e = exp(1LL * w * w % p, q / 2);
  if (q % 2 == 0) return e;
  return (1LL * e * w % p);
}
int euclid(int a, int b, int *x, int *y) {
  if (b == 0) {
    *x = 1;
    *y = 0;
    return a;
  }
  int x0, y0;
  int d = euclid(b, a % b, &x0, &y0);
  *x = y0;
  *y = x0 - (a / b) * y0;
  return d;
}
int inversmodular(int a) {
  int x, y;
  int d = euclid(a, p, &x, &y);
  return ((x + p) % p);
}
int main() {
  int i, j;
  scanf("%d %d %d %I64d", &a, &b, &p, &x);
  for (j = 1; j < p; j++) {
    y = j - 1LL * b * inversmodular(exp(a, j)) % p;
    y = (y + p) % p;
    if (x - j < 0) continue;
    z = (x - j) / (p - 1);
    z -= y;
    if (z >= 0) {
      sol += z / p + 1;
    }
  }
  printf("%I64d", sol);
  return 0;
}

```