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 long long maxn = 1e6 + 500;
long long A[maxn], A_ni[maxn], inv[maxn];
void pre(long long mod) {
  A[0] = A_ni[0] = A[1] = A_ni[1] = inv[1] = 1;
  for (int i = 2; i < maxn; i++) A[i] = A[i - 1] * i % mod;
  for (int i = 2; i < maxn; i++) inv[i] = (mod - mod / i) * inv[mod % i] % mod;
  for (int i = 2; i < maxn; i++) A_ni[i] = A_ni[i - 1] * inv[i] % mod;
}
long long C(long long n, long long m, long long mod) {
  return A[n] * A_ni[m] % mod * A_ni[n - m] % mod;
}
long long ksm(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
long long tim[maxn], sum[maxn];
int main() {
  long long n, m, a, q;
  cin >> n >> m >> a >> q;
  long long x = 1, now = a % q;
  while (now != 1) {
    now = now * a % q;
    x++;
  }
  pre(x);
  for (int i = 1; i <= m + 1; i++) {
    tim[i] = (tim[i - 1] + C(m, i - 1, x)) % x;
  }
  for (int i = n; i >= 1; i--) {
    int index = n - i + 1;
    if (index <= m + 1)
      sum[i] = ksm(a, tim[index], q);
    else
      sum[i] = sum[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    printf("%lld", sum[i]);
    if (i != n) printf(" ");
  }
  puts("");
  return 0;
}
