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 N = 1e6 + 200, D = 12;
int n, m, a, q, mod;
long long fac[N], inv[N], sum[N];
long long bp(long long a, long long p, long long mod) {
  long long r = 1;
  while (p > 0) {
    if (p & 1) r = r * a % mod;
    p >>= 1;
    a = a * a % mod;
  }
  return r;
}
long long c(int n, int k) { return fac[n] * inv[n - k] % mod * inv[k] % mod; }
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> a >> q;
  long long x = 1;
  for (int i = 1; i < N; i++) {
    x = x * a % q;
    if (x == 1) {
      mod = i;
      break;
    }
  }
  fac[0] = inv[0] = 1;
  for (int i = 1; i < N; i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = bp(fac[i], mod - 2, mod);
  }
  sum[0] = 1;
  for (int i = 1; i <= m; i++) {
    sum[i] = (sum[i - 1] + c(m, i)) % mod;
  }
  for (int i = 0; i < n; i++) {
    cout << bp(a, sum[min(m + 1, n - i) - 1], q) << " ";
  }
  return 0;
}
