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;
long long MOD;
long long MOD_p;
const int NMAX = 1000123;
long long comb[NMAX + 1];
long long MOD_pinv[NMAX + 1];
long long psum[NMAX + 1];
long long Bpow[NMAX + 1];
int main() {
  int N, M, B, Q;
  scanf("%d%d%d%d", &N, &M, &B, &Q);
  MOD = Q;
  MOD_p = 1;
  long long tmp = B;
  while (1) {
    tmp = tmp * B % Q;
    MOD_p++;
    if (tmp == 1) break;
  }
  MOD_pinv[1] = 1;
  for (int i = 2; i < N; i++)
    MOD_pinv[i] = (MOD_p - MOD_p / i) * MOD_pinv[MOD_p % i] % MOD_p;
  comb[0] = 1;
  for (int i = 1; i < N; i++) {
    if (i > M) continue;
    comb[i] = comb[i - 1] * (M - i + 1) % MOD_p;
    comb[i] = comb[i] * MOD_pinv[i] % MOD_p;
  }
  psum[0] = comb[0];
  for (int i = 1; i < N; i++) psum[i] = (psum[i - 1] + comb[i]) % MOD_p;
  Bpow[0] = 1;
  for (int i = 1; i <= NMAX; i++) Bpow[i] = Bpow[i - 1] * B % MOD;
  for (int i = 0; i < N; i++) {
    long long p = psum[N - i - 1];
    printf("%lld ", Bpow[p] % MOD);
  }
}
