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;
bool isPrime[1001005];
vector<long long> primes;
long long f[1001005], F[1001005];
long long modpow(long long a, long long b, long long m) {
  if (!b)
    return 1;
  else {
    long long z = modpow(a, b / 2, m) % m;
    z = (z * z) % m;
    if (b % 2)
      return (a * z) % m;
    else
      return z;
  }
}
int main() {
  long long N, M, a, Q;
  cin >> N >> M >> a >> Q;
  for (int i = 0; i < 1001005; i++) isPrime[i] = 1;
  for (int i = 2; i < 1001005; i++) {
    if (isPrime[i]) {
      primes.push_back(i);
      for (int k = i * 2; k < 1001005; k += i) isPrime[k] = 0;
    }
  }
  long long p = 0;
  for (int i = 0; i < primes.size(); i++) {
    if (modpow(a, primes[i], Q) == 1) {
      p = primes[i];
      break;
    }
  }
  f[0] = F[0] = 1;
  for (long long i = 1; i < 1001005; i++) {
    f[i] = (f[i - 1] % p * i % p) % p;
    F[i] = modpow(f[i], p - 2, p);
  }
  vector<long long> ans;
  long long add = 0;
  for (int i = 0; i <= N; i++) {
    long long t = f[M] % p;
    if (i <= M) {
      t = (t % p * F[i] % p) % p;
      t = (t % p * F[M - i] % p) % p;
      add = (add % p + t) % p;
    }
    ans.push_back(modpow(a, add % p, Q));
  }
  for (int i = N - 1; i >= 0; i--) {
    cout << ans[i];
    if (i) cout << " ";
  }
  cout << endl;
}
