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;
int m, n, a;
long long q, times = 1;
long long fact[1000125], inv_fact[1000125];
long long ans[1000125];
long long my_pow(long long a, long long m) {
  long long ret = 1;
  while (m) {
    if (m & 1) ret = (ret * a) % times;
    a = (a * a) % times, m >>= 1;
  }
  return ret;
}
void init() {
  long long pow_ans = a;
  while (pow_ans != 1ll) times++, pow_ans = (pow_ans * a) % q;
  fact[0] = 1ll, inv_fact[0] = 1ll;
  for (int i = 1; i <= m; i++)
    fact[i] = (fact[i - 1] * i) % times,
    inv_fact[i] = (inv_fact[i - 1] * my_pow(i, times - 2ll)) % times;
}
long long pow_q(long long a, long long m) {
  long long ret = 1;
  while (m) {
    if (m & 1) ret = (ret * a) % q;
    a = (a * a) % q, m >>= 1;
  }
  return ret;
}
void solve() {
  ans[n + 1] = 0;
  for (int i = n; i >= 1; i--) {
    ans[i] = ans[i + 1];
    if (n - i <= m) {
      long long first = (inv_fact[n - i] * inv_fact[m - (n - i)]) % times;
      ans[i] = (ans[i] + first * fact[m]) % times;
    }
  }
  for (int i = 1; i <= n; i++) ans[i] = pow_q(a, ans[i]);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
}
int main() {
  scanf("%d%d%d%lld", &n, &m, &a, &q);
  init();
  solve();
  return 0;
}
