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 long long MAXN = 1e6 + 123;
long long n, m, a, Q;
long long phi;
long long ans[MAXN];
pair<long long, long long> extgcd(long long a, long long b) {
  if (!b) return pair<long long, long long>(1, 0);
  pair<long long, long long> t = extgcd(b, a % b);
  return pair<long long, long long>(t.second, t.first - t.second * (a / b));
}
long long inv(long long a, long long p) {
  return (extgcd(a, p).first % p + p) % p;
}
long long pw(long long a, long long k) {
  long long t = 1;
  for (long long i = 1; i <= k; i <<= 1) {
    if (i & k) t = t * a % Q;
    a = a * a % Q;
  }
  return t;
}
signed main() {
  cin >> n >> m >> a >> Q;
  for (long long i = 1, j = a; i <= MAXN; i++, j = j * a % Q) {
    if (j == 1) {
      phi = i;
      break;
    }
  }
  ans[0] = 1;
  for (long long i = 1; i <= m; i++)
    ans[i] = (ans[i - 1] * (m - i + 1) % phi) * inv(i, phi) % phi;
  for (long long i = 1; i <= m; i++) ans[i] += ans[i - 1];
  for (long long i = m + 1; i < n; i++) ans[i] = ans[i - 1];
  for (long long i = 1; i <= n; i++) {
    cout << pw(a, ans[n - i]) << " \n"[i == n];
  }
}
