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 multiplicativeOrder(long long a, int Q) {
  int res = 1;
  for (int i = 1;; ++i) {
    res = res * a % Q;
    if (res == 1) return i;
  }
  return -1;
}
int power(long long a, int b, int MOD) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a % MOD;
    }
    a = a * a % MOD;
    b /= 2;
  }
  return res;
}
int main() {
  int M, a, Q, N;
  cin >> N >> M >> a >> Q;
  int k = multiplicativeOrder(a, Q);
  vector<int> pref(N, 0);
  pref[0] = 1;
  for (int i = 0; i + 1 < N; ++i) {
    pref[i + 1] = pref[i] * (long long)(M - i) * power(i + 1, k - 2, k) % k;
  }
  for (int i = 1; i < N; ++i) {
    pref[i] += pref[i - 1];
    if (pref[i] >= k) pref[i] -= k;
  }
  for (int i = N - 1; i >= 0; --i) {
    printf("%d ", power(a, pref[i], Q));
  }
  return 0;
}
