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, ph, p[1234567], f[1234567], c[1234567];
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &m, &a, &q);
  p[0] = 1, p[1] = a % q, ph = 1;
  while (p[ph] != 1) p[++ph] = p[ph - 1] * a % q;
  f[1] = 1;
  for (long long i = 2; i < ph; i++) f[i] = -(ph / i) * f[ph % i] % ph;
  _ = 1;
  c[0] = 1;
  for (long long i = 1; i < n; i++)
    _ = _ * (m - i + 1) % ph * f[i] % ph, c[i] = c[i - 1] + _, c[i] %= ph;
  for (long long i = 1; i <= n; i++)
    printf("%I64d%c", p[(c[n - i] + ph) % ph], i == n ? '\n' : ' ');
  return 0;
}
