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;
int n, m, a, q, i, pw[2000050], ni[2000050], c[2000050];
int main() {
  scanf("%d%d%d%d", &n, &m, &a, &q);
  int y = 1;
  pw[0] = 1;
  pw[1] = a;
  while (pw[y] != 1) y++, pw[y] = 1LL * pw[y - 1] * a % q;
  ni[1] = 1;
  for (i = 2; i < y; i++) ni[i] = -1LL * (y / i) * ni[y % i] % y;
  int x = 1;
  c[0] = 1;
  for (i = 1; i < n; i++)
    x = 1LL * x * (m - i + 1) % y * ni[i] % y, c[i] = (c[i - 1] + x) % y;
  for (i = 1; i <= n; i++) printf("%d ", pw[(c[n - i] + y) % y]);
  puts("");
  return 0;
}
