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;
constexpr int N = 2e6;
long long n, m, a, MOD, p[N], mul, ans[N], phi, fact[N];
long long mypow(long long a, long long b) {
  return b ? (mypow(a * a % MOD, b / 2) * (b & 1 ? a : 1)) % MOD : 1;
}
long long mypows(long long a, long long b) {
  return b ? (mypows(a * a % phi, b / 2) * (b & 1 ? a : 1)) % phi : 1;
}
long long choose(long long a, long long b) {
  if (b > a) return 0;
  return fact[a] * mypows((fact[b] * fact[a - b]) % phi, phi - 2);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m >> a >> MOD;
  p[0] = 1;
  fact[0] = 1;
  for (int i = 1; i < N; i++) p[i] = (p[i - 1] * a) % MOD;
  for (int i = 1; i < N; i++)
    if (p[i] == 1) {
      phi = i;
      break;
    }
  for (int i = 1; i < N; i++) fact[i] = (fact[i - 1] * i) % phi;
  for (int i = n - 1; i >= 0; i--) {
    mul += choose(m, n - 1 - i);
    mul %= phi;
    ans[i] = mypow(a, mul);
  }
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}
