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 p[2000100], sum[2000100];
int fac[2000100], rev[2000100];
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int main() {
  int N, M, a, P, Q;
  scanf("%d%d%d%d", &N, &M, &a, &Q);
  p[0] = 1;
  P = -1;
  for (int i = 1; i < 2000100; i++) {
    p[i] = 1LL * p[i - 1] * a % Q;
    if (p[i] == 1 && P < 0 && i > 1) P = i;
  }
  fac[0] = rev[0] = 1;
  while (P <= 1)
    ;
  for (int i = 1; i < 2000100; i++) {
    fac[i] = (long long)fac[i - 1] * i % P;
    rev[i] = power(fac[i], P - 2, P);
  }
  sum[0] = 1;
  for (int i = 1; i <= M; i++) {
    int c = (long long)fac[M] * rev[i] % P;
    c = (long long)c * rev[M - i] % P;
    sum[i] = ((long long)sum[i - 1] + c) % P;
  }
  for (int i = N - 1; i >= 0; i--) {
    int k = min(i, M);
    printf("%d ", p[sum[k]]);
  }
  return 0;
}
