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 d[1000234], b[1000234], c[1000234], ans[1000234], n, m, q, a, t, s;
bool f[1000234];
long long pow_(long long x, long long y, long long mod) {
  if (y == 0) return 1;
  long long tmp = pow_(x, y / 2, mod);
  tmp = tmp * tmp % mod;
  if (y % 2) tmp = tmp * x % mod;
  return tmp;
}
long long C(long long x, long long y) {
  if (y > x) return 0;
  if (y == 0 || y == x) return 1;
  return (b[x] * c[y] % t) * c[x - y] % t;
}
long long lucas(long long x, long long y) {
  if (y == 0) return 1;
  return (C(x % t, y % t) * lucas(x / t, y / t) % t);
}
int main() {
  scanf("%d%d%d%d", &n, &m, &a, &q);
  memset(f, true, sizeof(f));
  for (int i = 2; i < 1000234; i++)
    if (f[i])
      for (int j = 2; i * j < 1000234; j++) f[i * j] = false;
  t = 1;
  s = a % q;
  while (s % q != 1) {
    t++;
    s = s * a % q;
  }
  b[0] = 1;
  c[0] = 1;
  for (int i = 1; i < t; i++) {
    b[i] = b[i - 1] * i % t;
    c[i] = pow_(b[i], t - 2, t);
  }
  ans[0] = 1;
  for (int i = 1; i <= m; i++) {
    long long tt = lucas(m, i);
    ans[i] = (ans[i - 1] + tt) % t;
  }
  for (int i = m + 1; i < n; i++) ans[i] = ans[i - 1];
  for (int i = 1; i <= n; i++) printf("%d ", pow_(a, ans[n - i], q));
  printf("\n");
  return 0;
}
