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 inf = 1000000005;
const long long INF = 1000000000000000005LL;
const int MN = 200005, MM = 2000005;
long long mod;
long long dp[MM], A[MM];
long long a, Q;
long long fmp(long long a, long long b) {
  if (!b) return 1LL;
  long long hf = fmp(a, b >> 1);
  hf = (hf * hf) % mod;
  if (b & 1LL) hf = (hf * a) % mod;
  return hf;
}
long long revMod(long long x) { return fmp(x, mod - 2LL); }
long long findMod() {
  A[0] = 1;
  for (int i = 1; i < MM; ++i) {
    A[i] = (A[i - 1] * a) % Q;
    if (A[i] == 1) return (long long)i;
  }
}
int main() {
  int n, m;
  scanf("%d%d%lld%lld", &n, &m, &a, &Q);
  mod = findMod();
  long long cur = m % mod;
  dp[1] = a % Q;
  for (int i = 2; i <= min(n, m + 1); ++i) {
    dp[i] = (dp[i - 1] * A[(int)cur]) % Q;
    cur *= (long long)(m - i + 1);
    cur %= mod;
    cur *= revMod((long long)(i));
    cur %= mod;
  }
  for (int i = m + 2; i <= n; ++i) dp[i] = dp[i - 1];
  for (int i = n; i > 0; --i) printf("%lld ", dp[i]);
}
