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 = (int)1e6 + 100;
const int maxM = (int)1e6 + 100;
const int maxlog = (int)20;
const int P = 1e6 + 7;
int n, m, a, q, res[maxn];
int binpow(int a, int b, int mod) {
  if (b == 0) return 1 % mod;
  if (b % 2 == 0) {
    int d = binpow(a, b / 2, mod);
    return d * 1ll * d % mod;
  } else {
    int d = binpow(a, b - 1, mod);
    return d * 1ll * a % mod;
  }
}
int main() {
  scanf("%d%d%d%d", &n, &m, &a, &q);
  a %= q;
  int mod = 1, cur = a;
  while (cur != 1) {
    mod++;
    cur = (cur * 1ll * a) % q;
  }
  res[0] = 1;
  for (int i = 1; i < n && i <= m; i++) {
    res[i] = res[i - 1] * 1ll * (m - i + 1) % mod;
    res[i] = res[i] * 1ll * binpow(i, mod - 2, mod) % mod;
  }
  for (int i = 1; i < n; i++) res[i] = (res[i - 1] + res[i]) % mod;
  for (int i = n - 1; i >= 0; i--) printf("%d ", binpow(a, res[i], q));
}
