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 int MAXN = 1001000;
int nowarn;
int n, m, a, q, o;
int f[MAXN];
int fpow(int a, int b, int mod) {
  int c = 1;
  while (b) {
    if (b & 1) c = int((long long)c * a % mod);
    a = int((long long)a * a % mod);
    b >>= 1;
  }
  return c;
}
int inv(int a, int mod) { return fpow(a, mod - 2, mod); }
void prelude() {
  int phi = q, tmp = q;
  for (int i = 2; i * i <= tmp; ++i)
    if (tmp % i == 0) {
      phi = phi / i * (i - 1);
      while (tmp % i == 0) tmp /= i;
    }
  if (tmp > 1) phi = phi / tmp * (tmp - 1);
  o = 1e9;
  for (int i = 2; i * i <= phi; ++i)
    if (phi % i == 0) {
      if (fpow(a, i, q) == 1) o = min(o, i);
      if (fpow(a, phi / i, q) == 1) o = min(o, phi / i);
    }
}
int main() {
  nowarn = scanf("%d%d%d%d", &n, &m, &a, &q);
  prelude();
  memset(f, 0, sizeof(f));
  f[0] = 1;
  for (int i = 1; i < min(n, m + 1); ++i)
    f[i] = int((long long)f[i - 1] * (m - i + 1) % o * inv(i, o) % o);
  for (int i = 1; i < n; ++i) f[i] = (f[i] + f[i - 1]) % o;
  for (int i = n - 1; i >= 0; --i) printf("%d ", fpow(a, f[i], q));
  puts("");
  return 0;
}
