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;
long long Pow(long long x, long long k, long long p) {
  long long ans = 1;
  for (; k; k >>= 1, (x *= x) %= p)
    if (k & 1) (ans *= x) %= p;
  return ans;
}
long long n, m, a, q, fac[2010005], inv[2010005], ans[2010005], p, tt;
int main() {
  scanf("%lld%lld%lld%lld", &n, &m, &a, &q), a %= q;
  for (tt = a, p = 1; tt % q != 1; ++p, (tt *= a) %= q)
    ;
  fac[0] = 1;
  for (int i = (1); i <= (m + 10); ++i) fac[i] = fac[i - 1] * i % p;
  inv[1] = 1;
  for (int i = (2); i <= (m + 10); ++i)
    inv[i] = (long long)(p - p / i) * inv[p % i] % p;
  inv[0] = 1;
  for (int i = (1); i <= (m + 10); ++i) inv[i] = inv[i - 1] * inv[i] % p;
  for (int i = (0); i <= (min(m, n)); ++i)
    ans[i] = fac[m] * inv[i] % p * inv[m - i] % p;
  (ans[0] += p) %= p;
  for (int i = (1); i <= (n - 1); ++i)
    ans[i] = (ans[i] + ans[i - 1]) % p, (ans[i] += p) %= p;
  reverse(ans, ans + n);
  for (int i = (0); i <= (n - 1); ++i) ans[i] = Pow(a, ans[i], q);
  for (int i = (0); i <= (n - 1); ++i) printf("%lld ", ans[i]);
  return 0;
}
