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;
using ll = long long;
using db = double;
using vi = vector<int>;
const int maxn = 2000000;
ll n, m, a, P, Q;
ll inv[maxn], fac[maxn], ifac[maxn];
ll ans[maxn], q[maxn];
ll C(int a, int b) {
  if (a < b) return 0;
  return fac[a] * ifac[b] % P * ifac[a - b] % P;
}
int main() {
  cin >> n >> m >> a >> Q;
  q[0] = 1;
  for (int i = 1; i < maxn; ++i) {
    q[i] = q[i - 1] * a % Q;
    if (q[i] == 1) {
      P = i;
      break;
    }
  }
  inv[1] = fac[0] = fac[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i < maxn; ++i) {
    inv[i] = P / i * (P - inv[P % i]) % P;
    fac[i] = fac[i - 1] * i % P;
    ifac[i] = ifac[i - 1] * inv[i] % P;
  }
  ans[0] = 1;
  for (int i = 1; i < n; ++i) ans[i] = (ans[i - 1] + C(m, i)) % P;
  for (int i = n - 1; i >= 0; --i) printf("%lld ", q[ans[i]]);
  puts("");
}
