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>
const int N = 2e6;
using namespace std;
long long n, m, A, Q, X;
long long ans[N], inv[N], jiecheng[N], C[N];
void getx() {
  long long sum = A;
  long long num = 1;
  while (sum != 1) {
    sum = sum * A % Q;
    num++;
  }
  X = num;
}
long long quickpow(long long a, long long b, long long mod) {
  long long now = 1;
  while (b) {
    if (b & 1) now = now * a % mod;
    a = a * a % mod;
    b = b >> 1;
  }
  return now;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &m, &A, &Q);
  A = A % Q;
  getx();
  inv[X - 1] = X - 1;
  for (int i = X - 2; i >= 0; --i) inv[i] = inv[i + 1] * (i + 1) % X;
  jiecheng[0] = 1;
  for (long long i = 1; i <= X; ++i)
    jiecheng[i] = (long long)jiecheng[i - 1] * i % X;
  long long last = 0;
  for (long long i = 0; i <= min(n - 1, m); ++i) {
    C[i] = jiecheng[m] % X * inv[i] % X * inv[m - i] % X;
    last = (last + C[i]) % X;
    ans[n - i] = quickpow(A, last, Q);
  }
  if (m < n)
    for (int i = 1; i <= n - m - 1; ++i) ans[i] = ans[n - m];
  for (int i = 1; i <= n; ++i) printf("%I64d ", ans[i]);
  return 0;
}
