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 maxn = 1e6 + 1000;
int n, m, a, Q, ans[maxn];
int p;
int get_p(int x, int mod) {
  int i = x, s = 1;
  while (i != 1) i = 1LL * i * x % mod, s++;
  return s;
}
int Pow(int x, int k, int mod) {
  int s = 1;
  for (; k; k >>= 1, x = 1LL * x * x % mod)
    if (k & 1) s = 1LL * s * x % mod;
  return s;
}
int main() {
  int i, j;
  scanf("%d%d%d%d", &n, &m, &a, &Q);
  p = get_p(a, Q);
  for (ans[0] = 1, i = 1; i <= m; i++)
    ans[i] = 1LL * ans[i - 1] * (m - i + 1) % p * Pow(i, p - 2, p) % p;
  for (i = 1; i <= m; i++) ans[i] = (ans[i] + ans[i - 1]) % p;
  for (i = m + 1; i < n; i++) ans[i] = ans[i - 1];
  for (i = 0; i < n; i++) ans[i] = Pow(a, ans[i], Q);
  for (i = n - 1; i >= 0; i--) printf("%d%c", ans[i], i ? ' ' : '\n');
  return 0;
}
