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 MAXN = 2000005;
int N, M, A, Q, P;
long long fac[MAXN];
long long inv[MAXN];
long long power[MAXN];
long long ans[MAXN];
long long expo(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  if (b % 2 == 0) return expo(a * a % P, b / 2);
  return a * expo(a * a % P, (b - 1) / 2) % P;
}
long long C(int n, int r) {
  if (n < r) return 0;
  return fac[n] * (inv[n - r] * inv[r] % P) % P;
}
int getOrder() {
  power[0] = 1;
  power[1] = A;
  int ord = 0;
  for (int i = 2; i < MAXN; i++) {
    power[i] = power[i - 1] * A % Q;
    if (power[i] == 1 && ord == 0) ord = i;
  }
  return ord;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> N >> M >> A >> Q;
  P = getOrder();
  fac[0] = inv[0] = 1;
  for (int i = 1; i < MAXN; i++) {
    fac[i] = fac[i - 1] * i % P;
    inv[i] = expo(fac[i], P - 2);
  }
  long long cur = 0;
  for (int i = 0; i < N; i++) {
    cur = (cur + C(M, i)) % P;
    ans[N - 1 - i] = power[cur];
  }
  for (int i = 0; i < N; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
