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 + 222;
int n, m, a, mod;
long long c[MAXN], ans[MAXN], fac[MAXN], inv[MAXN];
long long qpow(long long a, long long n) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
void init(long long p) {
  int n = 1e6 + 200;
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= n; i++) {
    fac[i] = fac[i - 1] * i % p;
    inv[i] = (p - p / i) * inv[p % i] % p;
  }
  for (int i = 2; i <= n; i++) inv[i] = inv[i - 1] * inv[i] % p;
}
long long C(long long n, long long m, int p) {
  if (n < m) return 0;
  return fac[n] * inv[m] % p * inv[n - m] % p;
}
int main() {
  cin >> n >> m >> a >> mod;
  long long p = 1, now = a;
  while (now != 1) now = now * a % mod, ++p;
  init(p);
  for (int i = 1; i <= n; i++) {
    c[i] = C(m, i - 1, p);
    c[i] = (c[i] + c[i - 1]) % p;
  }
  for (int i = n; i >= 1; i--) {
    if (n - m - 1 >= i)
      ans[i] = ans[i + 1];
    else
      ans[i] = qpow(a, c[n - i + 1]);
  }
  for (int i = 1; i <= n; i++) printf("%I64d%c", ans[i], " \n"[i == n]);
  return 0;
}
