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 N, M, A, Q;
int paq;
int phi(int A, int Q) {
  int aa = 1;
  for (int i = 1;; ++i) {
    aa = 1LL * aa * A % Q;
    if (aa == 1) return i;
  }
}
int pow(int a, int b, int m) {
  if (b == 0) return 1;
  int ans = pow(a, b / 2, m);
  ans = (1LL * ans * ans) % m;
  if (b % 2 == 1) ans = (1LL * ans * a) % m;
  return ans;
}
int main() {
  scanf("%d%d%d%d", &N, &M, &A, &Q);
  paq = phi(A, Q);
  vector<int> fuck;
  long long int comb = 1;
  int a = 0;
  for (int i = 0; i < N; ++i) {
    a += comb;
    a %= paq;
    fuck.push_back(a);
    comb *= (1LL * (M - i) * pow(i + 1, paq - 2, paq)) % paq;
    comb %= paq;
  }
  for (int i = N - 1; i >= 0; --i) {
    printf("%d ", pow(A, fuck[i], Q));
  }
  puts("");
  return 0;
}
