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 = 107;
const int MAXN = 1e6 + 7;
long long n, m, a, q;
long long quick_mod(long long a, long long b, long long mod) {
  long long ans = 1, base = a % mod;
  while (b) {
    if (b & 1) ans = ans * base % mod;
    base = base * base % mod;
    b >>= 1;
  }
  return ans;
}
long long num[MAXN], two[MAXN];
void init(long long mod) {
  num[1] = 1;
  for (long long i = 2; i <= m; ++i) {
    num[i] = num[i - 1] * i % mod;
  }
  two[0] = 1;
  for (long long i = 1; i <= n; ++i) {
    two[i] = two[i - 1] * 2 % mod;
  }
}
long long C(long long n, long long m, long long mod) {
  long long ans = num[n];
  ans = ans * quick_mod(num[m], mod - 2, mod) % mod;
  ans = ans * quick_mod(num[n - m], mod - 2, mod) % mod;
  return ans;
}
long long ans[MAXN];
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &m, &a, &q);
  long long mod = 1;
  long long x = a;
  for (;; mod++) {
    x %= q;
    if (x == 1) break;
    x *= a;
  }
  init(mod);
  long long i;
  ans[n] = 1;
  for (i = n - 1; i >= 1; --i) {
    long long p = n - i;
    if (p == m) break;
    ans[i] = (ans[i + 1] + C(m, p, mod)) % mod;
  }
  for (; i >= 1; --i) ans[i] = two[m];
  for (int i = 1; i <= n; ++i) {
    ans[i] = quick_mod(a, ans[i], q);
  }
  for (int i = 1; i <= n; ++i) printf("%I64d ", ans[i]);
  return 0;
}
