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;
int N, M, a, Q;
int pw[1001000];
int X;
int fac[1001000], inv[1001000];
int ans[1001000];
int C(int n, int m) { return (long long)fac[n] * inv[m] % X * inv[n - m] % X; }
int main() {
  scanf("%d%d%d%d", &N, &M, &a, &Q);
  X = 1;
  pw[0] = 1, pw[1] = a;
  while (pw[X] != 1) pw[X + 1] = (long long)pw[X] * a % Q, ++X;
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = 2; i <= M; ++i) {
    fac[i] = (long long)fac[i - 1] * i % X;
    inv[i] = (long long)(X - X / i) * inv[X % i] % X;
  }
  for (int i = 2; i <= M; ++i) inv[i] = (long long)inv[i - 1] * inv[i] % X;
  int cnt = 0;
  for (int i = 0; i < N; ++i) {
    cnt += i <= M ? C(M, i) : 0;
    cnt %= X;
    ans[N - i - 1] = pw[cnt];
  }
  for (int i = 0; i < N; ++i) printf("%d ", ans[i]);
  return 0;
}
