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;
const double pi = acos(-1.0);
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
const int N = 1100000;
int n, m, a, Q, P;
int ans[N], jc[N], ny[N];
int c(int n, int k) { return 1ll * jc[n] * ny[k] % P * ny[n - k] % P; }
int qpow(int x, int times, int mod) {
  int now = 1;
  while (times) {
    if (times & 1) now = (long long)now * x % mod;
    x = (long long)x * x % mod;
    times >>= 1;
  }
  return now;
}
int main() {
  n = read();
  m = read();
  a = read();
  Q = read();
  int now = 1;
  for (int i = 1;; i++) {
    now = 1LL * now * a % Q;
    if (now == 1) {
      P = i;
      break;
    }
  }
  jc[0] = 1;
  for (int i = 1; i <= m; i++) jc[i] = 1LL * jc[i - 1] * i % P;
  ny[m] = qpow(jc[m], P - 2, P);
  for (int i = m - 1; i >= 0; i--) ny[i] = 1ll * ny[i + 1] * (i + 1) % P;
  for (int i = 1; i <= min(n, m + 1); i++)
    ans[n - i + 1] = (ans[n - i + 2] + c(m, i - 1)) % P;
  for (int i = n - min(n, m + 1); i >= 1; i--) ans[i] = ans[i + 1];
  for (int i = 1; i <= n; i++) printf("%d ", qpow(a, ans[i], Q));
  return 0;
}
