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;
long long f[(int)(2e6 + 10)], c[(int)(2e6 + 10)], ans[(int)(2e6 + 10)], i, j, k,
    n, m, a, x, q, q2;
long long fe(long long x, long long y, long long q) {
  if (y == 0) return 1;
  long long t = fe(x, y / 2, q);
  if (y & 1) return t * t % q * x % q;
  return t * t % q;
}
int main() {
  cin >> n >> m >> a >> q;
  x = a;
  for (i = 1; i <= (int)(2e6 + 10); i++) {
    x = x * a % q;
    if (x == 1) break;
  }
  q2 = i + 1;
  f[0] = 1;
  for (i = 1; i <= m; i++) f[i] = f[i - 1] * i % q2;
  for (i = 0; i <= m; i++)
    c[i] = f[m] * fe(f[i], q2 - 2, q2) % q2 * fe(f[m - i], q2 - 2, q2) % q2;
  for (i = 1; i <= n; i++) ans[i] = (c[i - 1] + ans[i - 1]) % q2;
  for (i = 1; i <= n; i++) cout << fe(a, ans[n - i + 1], q) << ' ';
}
