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 n, m, a, q, an[1001000], mod, jc[1001000] = {0}, ny[1001000] = {0},
                                        sc[1001000] = {0};
long long K(long long x, long long y) {
  long long z = 1;
  for (; y; y >>= 1) {
    if (y & 1) {
      z = (z * x) % mod;
    }
    x = (x * x) % mod;
  }
  return z;
}
long long C(long long x, long long y) {
  if (x < y) return 0;
  return jc[x] * ny[x - y] % mod * ny[y] % mod;
}
int main() {
  cin >> n >> m >> a >> q;
  an[0] = 1;
  for (an[mod = 1] = a; 1;) {
    a = (an[++mod] = a * an[1] % q);
    if (a == 1) break;
  }
  long long mx = min(1000000ll, mod - 1);
  jc[0] = 1;
  for (long long i = 1; i <= mx; i++) {
    jc[i] = jc[i - 1] * i % mod;
  }
  ny[mx] = K(jc[mx], mod - 2);
  for (long long i = mx - 1; i >= 0; i--) {
    ny[i] = (ny[i + 1] * (i + 1)) % mod;
  }
  for (long long i = 0; i <= n; i++) {
    sc[i] = C(m, i);
  }
  for (long long i = 1; i <= n; i++) {
    sc[i] = (sc[i - 1] + sc[i]) % mod;
  }
  for (long long i = 1; i <= n; i++) {
    printf("%I64d ", an[sc[n - i]]);
  }
}
