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;
long long n, m, a, q;
long long ans[1000000 + 200], inv[1000000 + 200];
long long C[1000000 + 200], mod = 1;
long long quick_mod(long long a, long long b, long long m) {
  long long ans = 1;
  a %= m;
  while (b) {
    if (b & 1) {
      ans = ans * a % m;
      b--;
    }
    b >>= 1;
    a = a * a % m;
  }
  return ans;
}
void init() {
  inv[1] = 1, C[0] = 1;
  for (int i = 2; i < mod; ++i) {
    inv[i] = ((mod / i) * inv[mod % i]) % mod;
    inv[i] = (-inv[i] + mod) % mod;
  }
  for (int i = 1; i <= m; i++) {
    C[i] = C[i - 1] * (m + 1 - i) % mod * inv[i] % mod;
  }
}
int main() {
  cin >> n >> m >> a >> q;
  long long temp = a;
  while (temp != 1) {
    temp = temp * a % q;
    mod++;
  }
  init();
  for (long long i = n; i >= (n - m + 1 > 1 ? n - m + 1 : 1); i--) {
    ans[i] = (ans[i + 1] + C[m - (n - i)]) % mod;
  }
  for (long long i = (n - m > 0 ? n - m : 0); i >= 1; i--) {
    if (i == n - m)
      ans[i] = (ans[i + 1] + 1) % mod;
    else
      ans[i] = ans[i + 1];
  }
  for (long long i = 1; i <= n; i++) {
    printf("%lld ", quick_mod(a, ans[i], q));
  }
}
