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;
int N, M, a, Q, mod, F[1010000], InvF[1010000], Res[1010000];
int Pow(int a, int b, int Mod) {
  int r = 1;
  while (b) {
    if (b & 1) r = 1ll * r * a % Mod;
    a = 1ll * a * a % Mod;
    b >>= 1;
  }
  return r;
}
int Comb(int a, int b) {
  return 1ll * F[a] * InvF[b] % mod * InvF[a - b] % mod;
}
int main() {
  scanf("%d%d%d%d", &N, &M, &a, &Q);
  int t = 1, i;
  for (i = 1;; i++) {
    t = 1ll * t * a % Q;
    if (t == 1) break;
  }
  mod = i;
  F[0] = 1;
  for (i = 1; i <= M; i++) F[i] = 1ll * F[i - 1] * i % mod;
  InvF[M] = Pow(F[M], mod - 2, mod);
  for (i = M - 1; i >= 0; i--) InvF[i] = 1ll * InvF[i + 1] * (i + 1) % mod;
  int s = 0;
  for (i = 0; i < N; i++) {
    if (i <= M) s = (s + Comb(M, min(M, i))) % mod;
    Res[N - i] = s;
  }
  for (i = 1; i <= N; i++) {
    printf("%d ", Pow(a, Res[i], Q));
  }
}
