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;
long long MOD = 1e9 + 7;
const long long N = 1e6 + 125;
long long arr[2 * N];
long long x[N];
long long bpow(long long a, long long b, long long m) {
  long long x = 1;
  long long y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % m;
      if (x < 0) x += m;
    }
    y = (y * y) % m;
    if (y < 0) y += m;
    b /= 2;
  }
  return x;
}
long long inv(long long a) { return (bpow(a, MOD - 2, MOD)); }
long long c(long long n, long long r) {
  long long nr = 1;
  for (long long i = 1; i < n + 1; i++) nr = (nr * i) % MOD;
  for (long long i = 1; i < n - r + 1; i++) nr = (nr * inv(i)) % MOD;
  for (long long i = 1; i < r + 1; i++) nr = (nr * inv(i)) % MOD;
  return nr;
}
int main() {
  ios::sync_with_stdio(false);
  long long n, m, a, q;
  cin >> n >> m >> a >> q;
  for (long long i = 1; i < N; i++) {
    if (bpow(a, i, q) == 1) {
      MOD = i;
      break;
    }
  }
  bool pr = false;
  for (long long i = 0; i < 2 * n - 1; i++) {
    if (m - n + i + 1 < 0)
      arr[i] = 0;
    else if (m - n + i + 1 > m)
      arr[i] = 0;
    else {
      if (!pr) {
        arr[i] = c(m, m - n + i + 1);
        pr = true;
      } else
        arr[i] = ((arr[i - 1] * (n - i)) % MOD * inv(m - n + i + 1)) % MOD;
    }
  }
  x[0] = 0;
  for (long long i = 0; i < n; i++) {
    x[0] = (x[0] + arr[i]) % MOD;
  }
  for (long long i = 1; i < n; i++) {
    x[i] = (x[i - 1] + arr[n + i - 1] + MOD - arr[i - 1]) % MOD;
  }
  for (long long i = 0; i < n; i++) {
    cout << bpow(a, x[i], q) << " ";
  }
  cout << "\n";
  return 0;
}
