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;
long long C[2111111], sum[2111111], ans[2111111];
long long Pow(long long a, long long n, long long mod) {
  if (n == 0) return 1;
  long long tmp = Pow(a, n / 2, mod);
  tmp = tmp * tmp % mod;
  if (n % 2) return a * tmp % mod;
  return tmp;
}
long long rev(long long a) { return Pow(a, mod - 2, mod); }
int main() {
  int N, M, a, Q;
  scanf("%d%d%d%d", &N, &M, &a, &Q);
  long long now = 1;
  for (int i = 1; i <= Q; i++) {
    now *= a;
    now %= Q;
    if (now == 1) {
      mod = i;
      break;
    }
  }
  for (int i = 0; i <= M; i++) {
    if (i == 0)
      C[i] = 1;
    else {
      C[i] = C[i - 1] * (M - i + 1) % mod * rev(i) % mod;
    }
    if (i == 0)
      sum[0] = C[i];
    else
      sum[i] = (sum[i - 1] + C[i]) % mod;
  }
  for (int i = 0; i < N; i++) {
    ans[i] = sum[min(M, N - 1 - i)];
    long long res = Pow(a, ans[i], Q);
    if (i) printf(" ");
    printf("%I64d", res);
  }
  puts("");
  return 0;
}
