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;
const int N = 1e6 + 124;
int ans[N];
long long P(int n, int p, int mod) {
  if (!p) return 1 % mod;
  if (p == 1) return n % mod;
  long long temp = P(n, p / 2, mod), m = (temp * temp) % mod;
  if (p & 1) m = (m * n) % mod;
  return m;
}
int main() {
  int n, m, a, mo, q;
  scanf("%d%d%d%d", &n, &m, &a, &q);
  for (int i = 1;; i++) {
    if (P(a, i, q) == 1) {
      mo = i;
      break;
    }
  }
  long long temp = 1, g = 1;
  ans[0] = P(a, 1 % mo, q);
  for (int i = 0; i < n - 1; i++) {
    temp = (temp * (m - i)) % mo;
    temp = (temp * P(i + 1, mo - 2, mo)) % mo;
    g = (g + temp) % mo;
    ans[i + 1] = P(a, g, q);
  }
  for (int i = n - 1; i >= 0; i--) {
    printf("%d ", ans[i]);
  }
  return 0;
}
