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 + 5000;
int mod, fact[N], c[N], phi, prc[N], pwra[N];
int rev(int x) {
  int pwr[40];
  pwr[0] = x;
  for (int i = 1; i < 35; i++) {
    pwr[i] = (pwr[i - 1] * 1ll * pwr[i - 1]) % mod;
  }
  int res = 1, v = phi - 1;
  for (int j = 30; j >= 0; j--) {
    int bit = (v >> j) & 1;
    if (bit) {
      res = (res * 1ll * pwr[j]) % mod;
    }
  }
  return res;
}
int getC(int n, int k) {
  int res = (fact[n] * 1ll * rev(fact[k])) % mod;
  res = (res * 1ll * rev(fact[n - k])) % mod;
  return res;
}
int main() {
  int n, m, a, q;
  scanf("%d %d %d %d", &n, &m, &a, &q);
  int rem = 1;
  pwra[0] = 1;
  for (int i = 1; i <= 1e6 + 500; i++) {
    rem = (rem * 1ll * a) % q;
    pwra[i] = rem;
    if (rem == 1) {
      mod = i;
      break;
    }
  }
  fact[0] = 1;
  for (int i = 1; i < N; i++) {
    fact[i] = (fact[i - 1] * 1ll * i) % mod;
  }
  phi = mod - 1;
  c[0] = 1;
  prc[0] = 1;
  for (int i = 1; i <= m; i++) {
    c[i] = getC(m, i);
    prc[i] = (prc[i - 1] + c[i]) % mod;
  }
  for (int i = 1; i <= n; i++) {
    if (i + m <= n) {
      printf("%d ", pwra[prc[m]]);
    } else {
      printf("%d ", pwra[prc[n - i]]);
    }
  }
  return 0;
}
