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 inf = 1000000000;
const long long longinf = 1000000000000000000;
const double PI = 3.141592653589793;
long long pw(long long a, long long p, long long mod) {
  long long ans = 1;
  while (p > 0) {
    if (p % 2 == 1) {
      ans = (ans * a) % mod;
    }
    a = (a * a) % mod;
    p /= 2;
  }
  return ans;
}
const int chk = 30000;
const int MAXM = 1001000;
long long fc[MAXM];
long long rfc[MAXM];
long long c(long long n, long long k, long long mod) {
  if (n < k) return 0;
  return ((fc[n] * rfc[k]) % mod) * rfc[n - k] % mod;
}
void solve() {
  long long n, m, a, q;
  cin >> n >> m >> a >> q;
  long long st = pw(a, chk, q);
  long long bs = a;
  long long phi = 1;
  long long q1 = q;
  for (int i = 2; i < 1000000; ++i) {
    if (q1 % i == 0) {
      phi *= i - 1;
      q1 /= i;
      while (q1 % i == 0) q1 /= i, phi *= i;
    }
  }
  if (q1 > 1) phi *= q1 - 1;
  long long md = phi;
  for (long long i = 1; i < 1000000; ++i) {
    if (phi % i == 0) {
      if (pw(a, i, q) == 1) {
        md = min(i, md);
      }
      if (pw(a, phi / i, q) == 1) {
        md = min(phi / i, md);
      }
    }
  }
  fc[0] = 1;
  for (int i = 1; i < md; ++i) {
    fc[i] = (fc[i - 1] * i) % md;
  }
  rfc[md - 1] = pw(fc[md - 1], md - 2, md);
  for (long long i = md - 2; i >= 0; --i) {
    rfc[i] = (rfc[i + 1] * (i + 1)) % md;
  }
  vector<long long> ans(n);
  ans[0] = 1;
  for (int i = 1; i < n; ++i) {
    ans[i] = c(m, i, md);
    ans[i] += ans[i - 1], ans[i] %= md;
  }
  for (int i = 0; i < n; ++i) {
    ans[i] = pw(a, ans[i], q);
  }
  reverse(ans.begin(), ans.end());
  for (int i = 0; i < n; ++i) {
    cout << ans[i] << ' ';
  }
  cout << endl;
}
int main() {
  cout.precision(20);
  cin.tie(0);
  iostream::sync_with_stdio(false);
  solve();
  return 0;
}
