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 = 2000126;
int n, m, a, q;
int factorials[MAXN];
int ans[MAXN];
int binpow(int a, int pwr, int q) {
  int res = 1;
  while (pwr > 0) {
    if (pwr & 1) res = (1ll * res * a) % q;
    a = (1ll * a * a) % q;
    pwr >>= 1;
  }
  return res;
}
int cnk(int n, int k, int phi) {
  if (k > n || k < 0) return 0;
  int nfact = factorials[n];
  int kfact = factorials[k];
  int nkfact = factorials[n - k];
  int res = (1ll * nfact * binpow(kfact, phi - 2, phi)) % phi;
  res = (1ll * res * binpow(nkfact, phi - 2, phi)) % phi;
  return res;
}
int main() {
  ios_base::sync_with_stdio();
  cin.tie(0);
  cin >> n >> m >> a >> q;
  int phi = 1;
  int cur = a;
  while (cur != 1) {
    cur = (1ll * cur * a) % q;
    phi++;
  }
  factorials[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    factorials[i] = (1ll * factorials[i - 1] * i) % phi;
  }
  ans[n] = 1;
  for (int i = n - 1; i >= 0; i--) {
    int power = cnk(m, n - 1 - i, phi);
    ans[i] = (1ll * ans[i + 1] * binpow(a, power, q)) % q;
  }
  for (int i = 0; i < n; i++) {
    cout << ans[i] << " ";
  }
}
