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;
inline int Read() {
  int x = 0, f = 1, c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  return x * f;
}
const int MAXN = 1100005;
int fac[MAXN], inv[MAXN], n, m, a, q, mod, pw[MAXN], f[MAXN];
inline int C(int x, int y) {
  if (x < y) return 0;
  return 1LL * fac[x] * inv[y] % mod * inv[x - y] % mod;
}
int main() {
  n = Read(), m = Read(), a = Read(), q = Read();
  pw[0] = fac[0] = inv[0] = fac[1] = inv[1] = 1;
  do {
    mod++;
    pw[mod] = 1LL * pw[mod - 1] * a % q;
  } while (pw[mod] ^ 1);
  for (int i = 2; i <= m; i++)
    fac[i] = 1LL * fac[i - 1] * i % mod,
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod;
  for (int i = 2; i <= m; i++) inv[i] = 1LL * inv[i] * inv[i - 1] % mod;
  for (int i = n; i; i--) f[i] = (f[i + 1] + C(m, n - i)) % mod;
  for (int i = 1; i <= n; i++) printf("%d%c", pw[f[i]], i == n ? '\n' : ' ');
  return 0;
}
