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 = 1e6 + 130;
long long fac[maxn];
long long inv[maxn];
long long c[maxn];
long long ans[maxn];
long long qpow(long long a, long long b, long long p) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return ans;
}
void init(int p) {
  fac[0] = 1;
  for (int i = 1; i <= maxn; i++) fac[i] = 1ll * fac[i - 1] * i % p;
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= maxn; i++) inv[i] = inv[p % i] * (p - p / i) % p;
  for (int i = 2; i <= maxn; i++) inv[i] = inv[i] * inv[i - 1] % p;
}
long long C(int n, int m, int p) {
  if (n == 0 || m == 0) return 1;
  return 1ll * fac[n] * inv[m] % p * inv[n - m] % p;
}
int main() {
  long long T, n, m, i, j, k, a, mod;
  cin >> n >> m >> a >> mod;
  int temp = 1;
  while (qpow(a, temp, mod) != 1) temp++;
  init(temp);
  for (int i = 1; i <= m + 1; i++) {
    c[i] = (1ll * C(m, i - 1, temp) + c[i - 1]) % temp;
  }
  for (int i = n; i; i--) {
    if (n - i + 1 <= m + 1)
      ans[i] = c[n - i + 1];
    else
      ans[i] = ans[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    printf("%lld", qpow(a, ans[i], mod));
    if (i < n)
      printf(" ");
    else
      printf("\n");
  }
}
