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 = 1e6 + 200;
long long arr[maxn];
long long pref[maxn];
long long modexp(long long a, long long ex, long long md) {
  long long ans = 1;
  while (ex > 0) {
    if (ex % 2 == 1) {
      ans = (ans * a) % md;
    }
    a = (a * a) % md;
    ex >>= 1;
  }
  return ans;
}
int main() {
  long long n, m, a, q;
  scanf("%lld", &n);
  scanf("%lld", &m);
  scanf("%lld", &a);
  scanf("%lld", &q);
  long long cur = a;
  long long p = 1;
  while (cur != 1) {
    cur = (cur * a) % q;
    p++;
  }
  pref[0] = 1;
  for (int i = 1; i <= m; i++) {
    pref[i] = ((pref[i - 1] * (m - i + 1)) % p) * modexp(i, p - 2, p);
    pref[i] %= p;
  }
  for (int i = 1; i <= m; i++) {
    pref[i] = (pref[i - 1] + pref[i]) % p;
  }
  for (int i = 1; i <= n; i++) {
    int z = min(m + 1, n - i + 1);
    printf("%lld", modexp(a, pref[z - 1], q));
    printf(" ");
  }
  printf("\n");
}
