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 n, m, a, q;
long long phiaq;
long long pmul[1000010];
long long pas[1000010];
long long ansp[1000010];
long long ap[1000010];
long long rmod(long long x, long long y = phiaq, long long ax = 1,
               long long ay = 0) {
  if (y == 0) {
    long long tmp = ax % phiaq;
    if (tmp < 0) tmp += phiaq;
    return tmp;
  }
  long long tk = x / y;
  return rmod(y, x % y, ay, ax - tk * ay);
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &m, &a, &q);
  {
    ap[0] = 1;
    long long now = a % q;
    for (phiaq = 1;; phiaq++) {
      ap[phiaq] = now;
      if (now == 1) break;
      now *= a;
      now %= q;
    }
  }
  pmul[0] = 1;
  for (int i = 1; i <= m + 1; i++) {
    pmul[i] = pmul[i - 1] * i;
    pmul[i] %= phiaq;
  }
  for (int i = 0; i < m + 1; i++) {
    pas[i] = pmul[m] * rmod(pmul[i]);
    pas[i] %= phiaq;
    pas[i] *= rmod(pmul[m - i]);
    pas[i] %= phiaq;
  }
  {
    long long now = 0;
    for (int i = 0; i < n; i++) {
      if (i <= m) now += pas[i];
      now %= phiaq;
      ansp[i] = now;
    }
  }
  for (int i = n - 1; i >= 0; i--) {
    printf("%lld ", ap[ansp[i]]);
  }
  printf("\n");
}
