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 N = 1100000;
int cj[N], inv[N];
int sum[N];
int power(int x, int y, int mod) {
  int ret = 1;
  for (; y; y >>= 1, x = 1ll * x * x % mod) {
    if (y & 1) ret = 1ll * ret * x % mod;
  }
  return ret;
}
int work(int a, int x, int mod) {
  int i;
  for (i = 2; i * i <= mod; i++) {
    if (x % i == 0) {
      if (power(a, i, mod) == 1) return i;
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) {
    if (power(a, x, mod) == 1) return x;
  }
  return mod;
}
int phi(int mod) {
  int i, x = mod;
  for (i = 2; i * i <= mod; i++) {
    if (x % i == 0) {
      mod = mod / i * (i - 1);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) mod = mod / x * (x - 1);
  return mod;
}
int main() {
  int n, m, a, p, mod;
  int i, j;
  scanf("%d%d%d%d", &n, &m, &a, &p);
  mod = work(a, phi(p), p);
  cj[0] = 1;
  for (i = 1; i <= m; i++) cj[i] = 1ll * cj[i - 1] * i % mod;
  inv[m] = power(cj[m], mod - 2, mod);
  inv[0] = 1;
  for (i = m - 1; i; i--) inv[i] = 1ll * inv[i + 1] * (i + 1) % mod;
  for (i = 0; i <= n; i++) {
    if (i > m) {
      sum[i] = 0;
      continue;
    }
    sum[i] = 1ll * cj[m] * inv[i] % mod * inv[m - i] % mod;
  }
  for (i = 1; i <= n; i++) sum[i] = (sum[i] + sum[i - 1]) % mod;
  for (i = 1; i <= n; i++) printf("%d ", power(a, sum[n - i], p));
}
