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 N = 2e6;
long long ans[N], fact[N], rev[N];
int phi;
int power(int a, int b, int mod) {
  if (b == 0) return 1;
  long long res = power(a, b / 2, mod);
  res = res * res % mod;
  if (b & 1) res *= a, res %= mod;
  return res;
}
long long comb(int n, int k) {
  if (k > n || k < 0) return 0;
  long long res = fact[n];
  res *= rev[k], res %= phi;
  res *= rev[n - k], res %= phi;
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k, a, mod;
  cin >> n >> k >> a >> mod;
  phi = 1;
  while (power(a, phi, mod) != 1) phi++;
  fact[0] = 1;
  for (int i = 1; i <= phi; i++) fact[i] = (1LL * fact[i - 1] * i) % phi;
  rev[phi - 1] = power(fact[phi - 1], phi - 2, phi);
  for (int i = phi - 2; i >= 0; i--)
    rev[i] = (1LL * rev[i + 1] * (i + 1)) % phi;
  long long ps = 0;
  for (int i = 0; i < n; i++) {
    ps = (ps + comb(k, i)) % phi;
    ans[i] = ps;
  }
  for (int i = n - 1; i >= 0; i--) {
    cout << power(a, ans[i], mod) << ' ';
  }
  cout << endl;
}
