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 su[2000005];
long long gopp[2000005];
int modd = 0;
long long divv(long long aa, long long bb) {
  long long lft = modd - 2;
  long long ans = 1;
  long long gop = bb;
  while (lft) {
    if (lft & 1) {
      ans = (ans * gop) % modd;
    }
    gop = (gop * gop) % modd;
    lft >>= 1;
  }
  ans = (ans * aa) % modd;
  return ans;
}
int main() {
  int n, m, a, q;
  scanf("%d%d%d%d", &n, &m, &a, &q);
  long long me = a;
  for (int i = 1;; i++) {
    if (me == 1) {
      modd = i;
      break;
    } else {
      me = (me * a) % q;
    }
  }
  su[n - 1] = 1;
  me = 1;
  for (int i = n - 2; i >= 0; i--) {
    me = (me * (m - n + 2 + i)) % modd;
    me = divv(me, n - 1 - i);
    su[i] = (su[i + 1] + me) % modd;
  }
  gopp[0] = 1;
  for (int i = 1; i <= modd; i++) {
    gopp[i] = (gopp[i - 1] * a) % q;
  }
  for (int i = 0; i < n; i++) {
    printf("%lld ", gopp[su[i] % modd]);
  }
  printf("\n");
}
