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 A[2000010], B[2000010], F[2000010], a, Q, M;
int n, m;
long long func(long long a, long long k, long long M) {
  long long s = 1;
  while (k > 0) {
    if (k & 1) s = s * a % M;
    k >>= 1;
    a = a * a % M;
  }
  return s;
}
long long C(int n, int m) {
  if (n < m || n < 0 || m < 0) return 0;
  if (n == m || m == 0) return 1;
  long long t = A[n] * B[m] % M;
  return t * B[n - m] % M;
}
int phi(long long a, long long Q) {
  long long s = 1;
  for (int i = 1;; i++) {
    s = s * a % Q;
    if (s == 1) return i;
  }
}
int main() {
  cin >> n >> m >> a >> Q;
  A[0] = B[0] = 1;
  M = phi(a, Q);
  for (int i = 1; i <= m; i++) A[i] = A[i - 1] * i % M;
  B[m] = func(A[m], M - 2, M);
  for (int i = m - 1; i >= 1; i--) B[i] = B[i + 1] * (i + 1) % M;
  for (int i = 0; i <= m; i++) F[i] = C(m, i);
  for (int i = 1; i <= n; i++) F[i] = (F[i] + F[i - 1]) % M;
  for (int i = 0; i <= n; i++) F[i] = func(a, F[i], Q);
  for (int i = n - 1; i >= 0; i--) printf("%I64d%c", F[i], i == 0 ? '\n' : ' ');
  return 0;
}
