Consider an array A with N elements, all being the same integer a.

Define the product transformation as a simultaneous update Ai = Ai·Ai + 1, that is multiplying each element to the element right to it for <image>, with the last number AN remaining the same. For example, if we start with an array A with a = 2 and N = 4, then after one product transformation A = [4, 4, 4, 2], and after two product transformations A = [16, 16, 8, 2].

Your simple task is to calculate the array A after M product transformations. Since the numbers can get quite big you should output them modulo Q.

Input

The first and only line of input contains four integers N, M, a, Q (7 ≤ Q ≤ 109 + 123, 2 ≤ a ≤ 106 + 123, <image>, <image> is prime), where <image> is the multiplicative order of the integer a modulo Q, see notes for definition.

Output

You should output the array A from left to right.

Example

Input

2 2 2 7


Output

1 2 

Note

The multiplicative order of a number a modulo Q <image>, is the smallest natural number x such that ax mod Q = 1. For example, <image>.
#include <bits/stdc++.h>
using namespace std;
long long m, n, q;
long long X;
long long f(long long a) {
  long long b = X - 2, c = X, k, ans = 1;
  while (b > 0) {
    long long p = 1, k = a;
    while (p * 2 <= b) {
      p *= 2;
      k = k * k % c;
    }
    ans = ans * k % c;
    b -= p;
  }
  return ans;
}
long long a[1005009], A;
long long p[1000509];
int main() {
  cin >> n >> m >> A >> q;
  p[0] = 1;
  for (long long i = 1;; i++) {
    p[i] = p[i - 1] * A % q;
    if (p[i] == 1) {
      X = i;
      break;
    }
  }
  a[0] = 1;
  long long x = m, t = 1;
  for (long long i = 1; i <= n - 1; i++) {
    a[i] = (a[i - 1] * x * f(i)) % X;
    x--;
  }
  long long s = 0;
  for (long long i = 0; i <= n - 1; i++) {
    s += a[i];
    a[i] = s;
  }
  for (long long i = n - 1; i >= 0; i--) cout << p[a[i] % X] << " ";
}
