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 n, m, a, q, p;
long long pa[1048576];
long long ans[1048576];
long long inv[1048576];
pair<int, int> egcd(int a, int b) {
  if (a == 0) return pair<int, int>(0, 1);
  pair<int, int> temp = egcd(b % a, a);
  return pair<int, int>((temp.second - b / a * temp.first) % p, temp.first);
}
int inver(int k) { return egcd(k, p).first; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m >> a >> q;
  pa[0] = 1;
  for (int i = 1; 1; ++i) {
    pa[i] = (pa[i - 1] * a) % q;
    if (pa[i] == 1) {
      p = i;
      break;
    }
  }
  for (int i = 0; i < p; ++i) {
    inv[i] = inver(i);
  }
  ans[0] = 1;
  long long bin = 1;
  for (int i = 1; i < n; ++i) {
    bin = bin * (m - i + 1) * inv[i] % p;
    ans[i] = ((ans[i - 1] + bin) % p + p) % p;
  }
  for (int i = n - 1; i >= 0; --i) {
    cout << pa[ans[i]];
    if (i) cout << " ";
  }
  cout << "\n";
  return 0;
}
