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 + 123;
long long n, m, a, q;
long long fi, p[maxn], c;
long long pow(long long x, long long y, long long z) {
  long long r = 1;
  for (; y > 0; y >>= 1) {
    if (y & 1) r = (r * x) % z;
    x = (x * x) % z;
  }
  return r;
}
int main() {
  cin >> n >> m >> a >> q;
  long long x = a % q;
  for (fi = 1; x != 1; fi++) x = (x * a) % q;
  p[0] = 1;
  c = 1;
  for (long long i = 1; i <= m; i++) {
    c = (c * (m - i + 1)) % fi;
    c = (c * pow(i, fi - 2, fi)) % fi;
    p[i] = (p[i - 1] + c) % fi;
  }
  for (int i = 0; i < n - m; i++) cout << pow(a, p[m], q) << " ";
  for (int i = min(m - 1, n - 1); i >= 0; i--) cout << pow(a, p[i], q) << " ";
  return 0;
}
