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 + 1;
long long n, m, a, mod, rev[N], fac[N], ans[N], P;
long long qpow(long long a, int b, long long mod) {
  long long ret = 1;
  while (b) {
    if (b & 1) ret = ret * a % mod;
    b >>= 1;
    a = a * a % mod;
  }
  return ret;
}
long long Lucas(long long n, long long k, int p) {
  long long ret = 1;
  while (n && k) {
    long long nn = n % p, kk = k % p;
    if (nn < kk) return 0;
    ret = ret * fac[nn] * qpow(fac[kk] * fac[nn - kk] % p, p - 2, p) % p;
    n /= p, k /= p;
  }
  return ret;
}
void pre() {
  fac[0] = 1;
  for (int i = 1; i <= m; ++i) fac[i] = fac[i - 1] * i % P;
  rev[m] = qpow(fac[m], P - 2, P);
  for (int i = m - 1; i >= 0; --i) rev[i] = rev[i + 1] * (i + 1) % P;
}
long long C(long long n, long long m) {
  return n < m ? 0 : fac[n] * rev[m] % P * rev[n - m] % P;
}
int main() {
  cin >> n >> m >> a >> mod;
  if (mod == 1) {
    for (int i = 1; i <= n; ++i) cout << 0 << " ";
    return 0;
  }
  long long now = 1;
  for (int i = 1;; ++i) {
    now = now * a % mod;
    if (now == 1) {
      P = i;
      break;
    }
  }
  pre();
  for (int i = 1; i <= n; ++i) ans[i] = C(m, i - 1);
  for (int i = 1; i <= n; ++i) (ans[i] += ans[i - 1]) %= P;
  for (int i = 1; i <= n; ++i) ans[i] = qpow(a, ans[i], mod);
  for (int i = n; i >= 1; --i) printf("%I64d ", ans[i]);
}
