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 ans[1111111], fac[1111111], inv[1111111], f;
long long quick(long long a, long long k, long long mod) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = ans * a % mod;
    a = a * a % mod;
    k /= 2;
  }
  return ans;
}
long long C(long long n, long long m) {
  return fac[n] * quick(fac[m], f - 2, f) * quick(fac[n - m], f - 2, f);
}
int main() {
  long long n, m, a, q, i, j, t;
  cin >> n >> m >> a >> q;
  t = a;
  for (f = 2;; f++) {
    t = (t * a) % q;
    if (t == 1) break;
  }
  fac[0] = 1;
  for (i = 1; i <= 1111110; i++) fac[i] = fac[i - 1] * i % f;
  ans[n] = 1;
  for (i = n - 1; i >= max(1LL, n - m); i--) {
    ans[i] = (ans[i + 1] + C(m, n - i)) % f;
  }
  for (; i >= 1; i--) ans[i] = ans[n - m];
  for (i = 1; i <= n; i++) printf("%lld ", quick(a, ans[i], q));
  printf("\n");
  return 0;
}
