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 = 1e6 + 1e3;
const long long INF = 1e18 + 18;
int n, m, a, Q;
int MOD;
int pw[maxn];
int gt[maxn], igt[maxn];
int ans[maxn];
void ReadInput() { cin >> n >> m >> a >> Q; }
int Pow(int first, int n) {
  if (n == 0) return 1;
  int t = Pow(first, n / 2);
  if (n % 2 == 0)
    return 1LL * t * t % MOD;
  else
    return 1LL * t * t % MOD * first % MOD;
}
void Init() {
  MOD = 1;
  pw[0] = 1;
  pw[1] = a;
  while (pw[MOD] != 1) {
    pw[MOD + 1] = 1LL * pw[MOD] * a % Q;
    ++MOD;
  }
  gt[0] = 1;
  for (int i = 1; i <= m; ++i) gt[i] = 1LL * gt[i - 1] * i % MOD;
  igt[m] = Pow(gt[m], MOD - 2);
  for (int i = m - 1; i >= 0; --i) igt[i] = 1LL * igt[i + 1] * (i + 1) % MOD;
}
int C(int k, int n) { return 1LL * gt[n] * igt[k] % MOD * igt[n - k] % MOD; }
void Solve() {
  int tot = 0;
  for (int i = 0; i < n; ++i) {
    tot = (1LL * tot + (i <= m ? C(i, m) : 0)) % MOD;
    ans[n - i - 1] = pw[tot];
  }
  for (int i = 0; i < n; ++i) cout << ans[i] << ' ';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  ReadInput();
  Init();
  Solve();
}
