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>
#pragma GCC optimize("inline")
using namespace std;
const int N = 1e6 + 200;
int A[N], B[N], x, n, m, a, q;
long long modulo(long long a, long long b, long long n) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y) % n;
    }
    y = (y * y) % n;
    b /= 2;
  }
  return x % n;
}
int main() {
  cin >> n >> m >> a >> q;
  A[0] = 1;
  long long p = n + 1;
  while (modulo(a, p, q) != 1) p++;
  for (long long i = 1; i < N; i++) {
    A[i] = (A[i - 1] * i) % p;
  }
  B[n - 1] = 1;
  for (int i = n - 2; i >= 0; i--) {
    long long f = A[m];
    if (n - i - 1 > m) {
      B[i] = B[i + 1];
    } else {
      long long k = A[n - i - 1], t = A[m - n + i + 1];
      k = (k * t) % p;
      k = modulo(k, p - 2, p);
      k = (k * f) % p;
      B[i] = (B[i + 1] + k) % p;
    }
  }
  for (int i = 0; i < n; i++) {
    cout << modulo(a, B[i], q) << " ";
  }
}
