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 = 1e6 + 1234;
int modx;
inline int sum(int a, int b, int mod) { return (a + b) % mod; }
inline int mult(int a, int b, int mod) { return (long long)a * b % mod; }
int n, m, e, q;
inline int binpow(int a, int b, int mod) {
  if (!b) {
    return 1;
  }
  if (b & 1) {
    return mult(a, binpow(a, b ^ 1, mod), mod);
  }
  return binpow(mult(a, a, mod), b >> 1, mod);
}
int fact[N], invfact[N];
void read() { cin >> n >> m >> e >> q; }
void precalc() {
  for (int x = 1;; ++x) {
    if (binpow(e, x, q) == 1) {
      modx = x;
      break;
    }
  }
  fact[0] = invfact[0] = 1;
  for (int i = 1; i < N; ++i) {
    fact[i] = mult(fact[i - 1], i, modx);
    invfact[i] = mult(invfact[i - 1], binpow(i, modx - 2, modx), modx);
  }
}
int c(int u, int v) {
  if (u > v) {
    return 0;
  }
  return mult(mult(invfact[u], invfact[v - u], modx), fact[v], modx);
}
int pw[N];
void solve() {
  read();
  precalc();
  pw[n] = 1;
  for (int i = n - 1; i > 0; --i) {
    pw[i] = sum(pw[i + 1], c(n - i, m), modx);
  }
  for (int i = 1; i <= n; ++i) {
    cout << binpow(e, pw[i], q) << " ";
  }
}
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0), cin.tie(0);
  solve();
  return 0;
}
