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;
int n, m, a, mod, i;
long long f[1000200], rev[1000200], sum[1000200], c[1000200];
long long POW(int a, int n) {
  if (!n) return 1;
  if (n & 1) return (a * POW(a, n - 1)) % mod;
  long long res = POW(a, n / 2);
  return (res * res) % mod;
}
void gcd(int a, int b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return;
  }
  long long x1, y1;
  gcd(b, a % b, x1, y1);
  x = y1;
  y = x1 - (a / b) * y1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cin >> n >> m >> a >> mod;
  long long x = 1, pw = a;
  for (; pw != 1; x++) pw = (pw * a) % mod;
  f[0] = rev[0] = 1;
  for (i = 1; i <= m; i++) {
    f[i] = (f[i - 1] * i) % x;
    long long y = f[i];
    gcd(f[i], x, rev[i], y);
    if (rev[i] < 0) rev[i] = (rev[i] + x * x) % x;
  }
  for (i = 0; i <= m; i++) {
    c[i] = (f[m] * ((rev[i] * rev[m - i]) % x)) % x;
    if (i) sum[i] = sum[i - 1];
    sum[i] = (sum[i] + c[i]) % x;
  }
  for (i = 0; i < n; i++) cout << POW(a, sum[min(n - i - 1, m)]) % mod << " ";
  return 0;
}
