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;
const int maxN = 1000000 + 300;
int N, M, a, Q, X;
int CM[maxN], F[maxN];
void findX() {
  int R = a % Q;
  X = 1;
  while (R != 1) {
    R = (1LL * R * a) % Q;
    X++;
  }
}
int poww(int x, int y, int MOD) {
  int R = 1;
  while (y > 0) {
    if (y & 1) R = (1LL * R * x) % MOD;
    x = (1LL * x * x) % MOD;
    y /= 2;
  }
  return R;
}
void findCM() {
  CM[0] = 1;
  for (int i = 1; i <= M; i++) {
    CM[i] = (1LL * CM[i - 1] * (M - i + 1)) % X;
    CM[i] = (1LL * CM[i] * poww(i, X - 2, X)) % X;
  }
}
void findF() {
  F[0] = 1;
  F[1] = CM[1];
  for (int i = 2; i < N; i++) {
    F[i] = (CM[i] + F[i - 2]) % X;
  }
}
void findR() {
  for (int i = N - 1; i >= 0; i--) {
    printf("%d", poww(a, F[i], Q));
    if (i == 0)
      printf("\n");
    else
      printf(" ");
  }
}
int main() {
  scanf("%d%d%d%d", &N, &M, &a, &Q);
  M++;
  findX();
  findCM();
  findF();
  findR();
  return 0;
}
