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 pwr(long long a, int x, int p) {
  if (!x) return 1;
  if (x == 1) return a;
  long long tmp = pwr(a, x >> 1, p);
  tmp = tmp * tmp % p;
  if (x & 1) tmp = tmp * a % p;
  return tmp;
}
long long frac[1001000], inv[1001000];
long long res[1001000];
int main(void) {
  int n, m, a, q;
  scanf("%d%d%d%d", &n, &m, &a, &q);
  long long pr = a % q;
  int pwrr = 1;
  while (pr != 1) {
    pr *= a;
    pr %= q;
    pwrr++;
  }
  long long base = pwr(2, m, pwrr);
  frac[0] = inv[0] = 1;
  frac[1] = 1;
  inv[1] = 1;
  for (int i = 2; i <= m; i++) {
    frac[i] = frac[i - 1] * i % pwrr;
    inv[i] = inv[i - 1] * pwr(i, pwrr - 2, pwrr) % pwrr;
  }
  for (int i = 1; i <= n; i++) {
    res[i] = base;
  }
  long long pre = 0;
  for (int i = 0; i <= m; i++) {
    pre += frac[m] * inv[i - 1] % pwrr * inv[m - i + 1] % pwrr;
    pre %= pwrr;
    if (n - m + i >= 1) {
      res[n - m + i] -= pre - pwrr;
      res[n - m + i] %= pwrr;
    }
  }
  for (int i = 1; i <= n; i++) {
    printf("%d ", pwr(a, res[i], q));
  }
  return 0;
}
