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 MX = 1111111;
int mod = 1000000007;
long long ans[MX], fact[MX];
long long inv[MX], f[MX], g[MX];
void init() {
  int i;
  inv[1] = 1;
  for (i = 2; i < MX; i++) inv[i] = inv[mod % i] * (mod - mod / i) % mod;
  f[0] = g[0] = 1;
  for (i = 1; i < MX; i++) {
    f[i] = f[i - 1] * i % mod;
    g[i] = g[i - 1] * inv[i] % mod;
  }
}
inline long long nCm(int n, int m) {
  if (m < 0 or m > n) return 0;
  return f[n] * g[m] % mod * g[n - m] % mod;
}
int main() {
  int x, n, m, a, q;
  scanf("%d%d%d%d", &n, &m, &a, &q);
  int p = a;
  fact[0] = 1;
  for (int i = 1; i < MX; i++) {
    fact[i] = 1LL * fact[i - 1] * a % q;
  }
  for (x = 1;; x++) {
    if (p == 1) break;
    p = 1LL * p * a % q;
  }
  mod = x;
  init();
  for (int i = n - 1; i >= 0; i--) ans[i] = nCm(m, n - 1 - i);
  ans[n - 1] = 1;
  for (int i = n - 2; i >= 0; i--) ans[i] = (ans[i + 1] + ans[i]) % mod;
  for (int i = 0; i < n; i++) {
    printf("%d%c", (int)fact[ans[i]], i < (n - 1) ? ' ' : '\n');
  }
  return 0;
}
