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 = 2e6 + 5;
int MOD;
long long power(long long a, long long n, long long m) {
  if (n == 0) return 1;
  long long p = power(a, n >> 1, m);
  p = (p * p) % m;
  if (n & 1) p = (p * a) % m;
  return p;
}
long long pp[N];
vector<long long> ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, a, q;
  cin >> n >> m >> a >> q;
  pp[0] = 1;
  for (int i = 1;; i++) {
    pp[i] = (pp[i - 1] * a) % q;
    if (pp[i] == 1) {
      MOD = i;
      break;
    }
  }
  long long sum = 0;
  long long cnt = 1;
  for (int i = 0; i < n; i++) {
    sum = (sum + cnt) % MOD;
    ans.push_back(pp[sum]);
    cnt = (cnt * (m - i)) % MOD;
    cnt = (cnt * power(i + 1, MOD - 2, MOD)) % MOD;
  }
  reverse(ans.begin(), ans.end());
  for (auto it : ans) {
    cout << it << " ";
  }
  cout << "\n";
  return 0;
}
