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 N = 1010101;
int n, m, a, q;
long long ans[N];
long long kpow(long long a, long long b, long long P) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= a) %= P;
    (a *= a) %= P;
    b >>= 1;
  }
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  cin >> n >> m >> a >> q;
  int x = 2;
  while (kpow(a, x, q) != 1) ++x;
  long long c = 1;
  for (int i = n; i; --i) {
    ans[i] = (ans[i + 1] + c) % x;
    int t = i - n + m;
    if (t < 0)
      c = 0;
    else
      c = c * t % x * kpow(m - t + 1, x - 2, x) % x;
  }
  for (int i = (1); i < (n + 1); i++) cout << kpow(a, ans[i], q) << " ";
  cout << "\n";
  return 0;
}
