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 = 2000000, mod = 1e9 + 7;
int w[MAXN];
stack<int> s;
long long pot(long long co, long long wyk, long long mod) {
  long long w = 1;
  for (; wyk; wyk /= 2) {
    if (wyk % 2) w = (w * co) % mod;
    co = (co * co) % mod;
  }
  return w;
}
int main() {
  int a, b, c, d, e, n, k, m, q, ord, sum;
  scanf("%d%d%d%d", &n, &m, &a, &q);
  w[0] = 1;
  b = a % q;
  for (ord = 1; b != 1; ord++) {
    w[ord] = b;
    b = ((long long)b * a) % q;
  }
  b = m;
  c = 1;
  d = 1;
  sum = 0;
  for (; c <= n; c++) {
    sum += d;
    sum %= ord;
    s.push(w[sum]);
    d = (long long)d * b * pot(c, ord - 2, ord) % ord;
    b--;
  }
  for (; !s.empty(); s.pop()) printf("%d ", s.top());
  return 0;
}
