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 = 1e6 + 1e5;
long long n, m, a, q, mod;
long long tmp = 1;
long long fac[maxn], inv_fac[maxn], inv[maxn], ans[maxn];
long long fpow(long long a, long long k) {
  int ans = 1;
  while (k) {
    if (k & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    k >>= 1;
  }
  return ans;
}
long long fpow(long long a, long long k, long long mod) {
  int ans = 1;
  while (k) {
    if (k & 1) ans = 1ll * ans * a % mod;
    a = 1ll * a * a % mod;
    k >>= 1;
  }
  return ans;
}
void init() {
  fac[0] = 1;
  for (int i = 1; i < maxn && i < mod; i++) fac[i] = 1ll * fac[i - 1] * i % mod;
  int st = min(1ll * maxn - 1, mod - 1);
  inv_fac[st] = fpow(fac[st], mod - 2);
  for (int i = st - 1; i >= 1; i--) {
    inv_fac[i] = 1ll * inv_fac[i + 1] * (i + 1) % mod;
  }
  inv_fac[0] = 1;
}
long long C(long long n, long long m) {
  return fac[n] * inv_fac[n - m] % mod * inv_fac[m] % mod;
}
int main() {
  cin >> n >> m >> a >> q;
  for (mod = 1; mod <= q; mod++) {
    tmp = tmp * a % q;
    if (tmp == 1) break;
  }
  init();
  ans[n] = 1;
  for (int i = n - 1; i >= 1; i--) {
    if (n - i >= m + 1)
      ans[i] = ans[i + 1];
    else
      ans[i] = (ans[i + 1] + C(m, n - i)) % mod;
  }
  for (int i = 1; i <= n; i++) cout << fpow(a, ans[i], q) << " ";
  return 0;
}
