#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000000 + 10;
int a, b, p, num[maxn];
long long xx, ans, X;
long long calc(long long x, long long y, long long md) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % md;
    y >>= 1;
    x = (x * x) % md;
  }
  return res;
}
int main() {
  scanf("%d%d%d%lld", &a, &b, &p, &X);
  long long val = 1;
  for (int i = 1; i < p; i++) {
    val = val * a % p;
    long long x = calc(val, p - 2, p) * b % p;
    long long MOD = (long long)p * (p - 1);
    long long x1 = (long long)p * i % MOD;
    long long x2 = (long long)(p - 1) * (p - 1) % MOD;
    x2 = x2 * x % MOD;
    long long res = (x1 + x2) % MOD;
    if (res == 0) res += MOD;
    ans += X / MOD;
    if (X % MOD >= res) ans++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Your task is to create 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 maxn = 1000000 + 10;
int a, b, p, num[maxn];
long long xx, ans, X;
long long calc(long long x, long long y, long long md) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % md;
    y >>= 1;
    x = (x * x) % md;
  }
  return res;
}
int main() {
  scanf("%d%d%d%lld", &a, &b, &p, &X);
  long long val = 1;
  for (int i = 1; i < p; i++) {
    val = val * a % p;
    long long x = calc(val, p - 2, p) * b % p;
    long long MOD = (long long)p * (p - 1);
    long long x1 = (long long)p * i % MOD;
    long long x2 = (long long)(p - 1) * (p - 1) % MOD;
    x2 = x2 * x % MOD;
    long long res = (x1 + x2) % MOD;
    if (res == 0) res += MOD;
    ans += X / MOD;
    if (X % MOD >= res) ans++;
  }
  cout << ans << endl;
  return 0;
}

```