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 + 10;
int Pow(int x, int t, int P) {
  int r = 1;
  for (; t; t >>= 1, x = 1ll * x * x % P)
    if (t & 1) r = 1ll * r * x % P;
  return r;
}
int x, n, m, a, Q, fac[N], ifac[N];
int comb(int a, int b) { return 1ll * fac[a] * ifac[a - b] % x * ifac[b] % x; }
int main() {
  scanf("%d%d%d%d", &n, &m, &a, &Q);
  x = 1;
  while (Pow(a, x, Q) != 1) ++x;
  fac[0] = 1;
  for (int i = 1; i < (x); ++i) fac[i] = 1ll * fac[i - 1] * i % x;
  ifac[x - 1] = Pow(fac[x - 1], x - 2, x);
  for (int i = (x - 1) - 1; i >= (0); --i)
    ifac[i] = 1ll * ifac[i + 1] * (i + 1) % x;
  vector<int> ans;
  int c = 1;
  for (int i = 0; i < (min(m, n)); ++i) {
    ans.push_back(Pow(a, c, Q));
    c = (c + comb(m, i + 1)) % x;
  }
  int t = Pow(a, Pow(2, m, x), Q);
  while ((int)(ans).size() < n) ans.push_back(t);
  reverse((ans).begin(), (ans).end());
  for (int i = 0; i < ((int)(ans).size()); ++i)
    printf("%d%c", ans[i], " \n"[i + 1 == (int)(ans).size()]);
  return 0;
}
