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>
long long n, m, a, Q, p;
long long inv[1001234];
long long comb[1001234];
long long ans[1001234];
long long mod[1001234];
int main() {
  int i, j;
  long long tmp;
  scanf("%lld%lld%lld%lld", &n, &m, &a, &Q);
  a %= Q;
  tmp = a;
  mod[0] = 1;
  for (p = 1; p <= Q; ++p) {
    mod[p] = tmp;
    if (tmp == 1) break;
    tmp *= a;
    tmp %= Q;
  }
  inv[1] = 1;
  for (i = 2; i < p; ++i) {
    inv[i] = ((p / i) * inv[p % i]) % p;
    inv[i] = (-inv[i] + p) % p;
  }
  comb[0] = 1;
  comb[1] = m % p;
  for (i = 2; i <= m; ++i)
    comb[i] = (((comb[i - 1] * (m - i + 1)) % p) * inv[i]) % p;
  ans[n] = 1;
  int c = 0;
  for (i = n - 1; i >= 1; --i) {
    ++c;
    if (c < m + 1) {
      ans[i] = ans[i + 1] + comb[c];
      ans[i] %= p;
    } else
      ans[i] = ans[i + 1];
  }
  for (i = 1; i <= n; ++i) ans[i] = mod[ans[i]];
  for (i = 1; i <= n; ++i) printf("%lld ", ans[i]);
}
