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 + 10;
int n, m, a, mod, p, ans[MAXN];
void get_p(int x) {
  int c = x, tot = 1;
  while (c != 1) {
    c = (1ll * c * x) % mod;
    ++tot;
  }
  p = tot;
}
int fpow(int x, int k, int mo) {
  int ret = 1;
  for (; k; k >>= 1, x = (1ll * x * x) % mo)
    if (k & 1) ret = (1ll * ret * x) % mo;
  return ret;
}
int main() {
  scanf("%d%d%d%d", &n, &m, &a, &mod);
  get_p(a);
  ans[0] = 1;
  for (int i = 1; i <= m; ++i) {
    int tmp = (1ll * ans[i - 1] * (m - i + 1)) % p;
    ans[i] = (1ll * tmp * fpow(i, p - 2, p)) % p;
  }
  for (int i = 1; i <= m; ++i) {
    ans[i] += ans[i - 1];
    ans[i] %= p;
  }
  for (int i = m + 1; i < n; ++i) ans[i] = ans[i - 1];
  for (int i = 0; i < n; ++i) ans[i] = fpow(a, ans[i], mod);
  for (int i = n - 1; i >= 0; --i) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}
