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 N = 1010000;
long long binpow(long long a, long long b, long long c) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % c;
    a = (a * a) % c;
    b >>= 1;
  }
  return res;
}
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
int phi;
int lp[N], pr[N], k = 0;
int cnt[N];
int n, m, mod, x;
long long go[N];
vector<pair<int, int> > fact(int x) {
  vector<pair<int, int> > res;
  while (x > 1) {
    if (res.size() > 0 && res[(int)res.size() - 1].first == lp[x])
      res[(int)res.size() - 1].second++;
    else
      res.push_back(make_pair(lp[x], 1));
    x /= lp[x];
  }
  return res;
}
long long f[N], inv[N];
long long C(int n, int m) {
  if (n < m) return 0;
  long long res = f[n];
  res = (res * inv[m]) % phi;
  res = (res * inv[n - m]) % phi;
  return res;
}
long long ans[N];
int main() {
  lp[0] = lp[1] = 1;
  for (int i = 2; i <= 1000123; ++i) {
    if (!lp[i]) {
      lp[i] = i;
      pr[++k] = i;
    }
    for (int j = 1; j <= k && pr[j] <= lp[i]; ++j) {
      if ((long long)pr[j] * i > 1000123) break;
      lp[pr[j] * i] = pr[j];
    }
  }
  cin >> n >> m >> x >> mod;
  for (int i = 1; i < mod; ++i) {
    if (binpow(x, i, mod) == 1) {
      phi = i;
      break;
    }
  }
  f[0] = inv[0] = 1;
  for (int i = 1; i <= 1000123; ++i) {
    f[i] = (f[i - 1] * i) % phi;
    inv[i] = binpow(f[i], phi - 2, phi);
  }
  go[0] = 1;
  for (int i = 1; i <= n; ++i) {
    go[i] = (go[i - 1] + C(m, i)) % phi;
  }
  for (int i = 1; i <= n; ++i) {
    ans[i] = binpow(x, go[i - 1], mod);
  }
  for (int i = n; i > 0; --i) printf("%lld ", ans[i]);
  return 0;
}
