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;
template <typename T, typename U>
inline void chkmax(T& x, U y) {
  if (x < y) x = y;
}
template <typename T, typename U>
inline void chkmin(T& x, U y) {
  if (y < x) x = y;
}
const int MAXN = 1100000;
int p[MAXN], C[MAXN], r[MAXN];
int main() {
  int N, M, A, Q;
  cin >> N >> M >> A >> Q;
  int P = 0;
  for (int i = 0, cur = 1; P == 0; cur = 1LL * cur * A % Q, i++) {
    p[i] = cur;
    if (i > 0 && cur == 1) P = i;
  }
  r[1] = 1;
  for (int i = 2; i <= M; i++) r[i] = (long long)(P - r[P % i]) * (P / i) % P;
  for (int i = 0, cur = 1; i <= M; i++)
    C[i] = ((i == 0 ? 0 : C[i - 1]) + cur) % P,
    cur = 1LL * cur * (M - i) % P * r[i + 1] % P;
  for (int i = 0; i < N; i++) printf("%d ", p[C[min(M, N - 1 - i)]]);
  puts("");
  return 0;
}
