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 modpow(long long a, long long n, long long m) {
  long long res;
  for (res = 1; n > 0; n >>= 1) {
    if (n & 1) res = (res * a) % m;
    a = (a * a) % m;
  }
  return res;
}
const int N = 1e6 + 124;
long long ans[N], pw[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  long long m, a, q;
  cin >> n >> m >> a >> q;
  int phi = 0;
  pw[0] = 1;
  for (int i = 1; phi == 0; i++) {
    pw[i] = (pw[i - 1] * a) % q;
    if (pw[i] == 1) phi = i;
  }
  long long v = 1, r = 0;
  for (int i = 0; i < n; i++) {
    r = (r + v) % phi;
    ans[i] = modpow(a, r, q);
    v = (v * (m - i)) % phi;
    v = (v * modpow(i + 1, phi - 2, phi)) % phi;
  }
  for (int i = n - 1; i >= 0; i--) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}
