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 inf = 1e6 + 123;
long long f[inf + 10], inv[inf + 10];
long long pa[inf + 10];
long long b[inf + 10];
int main() {
  int n, m, a, q, i;
  scanf("%d%d%d%d", &n, &m, &a, &q);
  int mod;
  long long res = 1;
  pa[0] = 1;
  for (i = 1; i <= inf; i++) {
    res = (res * a) % q;
    pa[i] = res;
    if (res == 1) break;
  }
  mod = i;
  assert(mod <= inf);
  f[0] = 1;
  for (i = 1; i <= m; i++) f[i] = (f[i - 1] * i) % mod;
  inv[0] = 1;
  inv[1] = 1;
  for (i = 2; i < mod; i++) inv[i] = mod - ((mod / i) * inv[mod % i]) % mod;
  for (i = 2; i < mod; i++) inv[i] = (inv[i] * inv[i - 1]) % mod;
  for (i = 0; i < min(n, m + 1); i++)
    b[i] = (((f[m] * inv[i]) % mod) * inv[m - i]) % mod;
  for (i = 1; i < n; i++) b[i] = (b[i] + b[i - 1]) % mod;
  for (i = n - 1; i >= 0; i--) printf("%I64d ", pa[b[i]]);
  return 0;
}
