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 n, m, a;
long long fac[1000125], inv[1000125], ans[1000125];
long long quickpow(long long p, long long k, long long mod) {
  long long ans = 1;
  while (k) {
    if (k & 1) ans = (ans * p) % mod;
    p = (p * p) % mod;
    k >>= 1;
  }
  return ans;
}
long long c(long long x, long long y, long long mod) {
  if (x < y) return 0;
  if (x == y || y == 0) return 1;
  return (fac[x] * inv[y] % mod * inv[x - y]) % mod;
}
int main() {
  long long mod, temp = 1, mo;
  scanf("%I64d%I64d%I64d%I64d", &n, &m, &a, &mod);
  for (int i = 1; i; i++) {
    temp = (temp * a) % mod;
    if (temp == 1) {
      mo = i;
      break;
    }
  }
  fac[0] = 1;
  for (int i = 1; i <= m; i++) {
    fac[i] = (fac[i - 1] * i) % mo;
    inv[i] = quickpow(fac[i], mo - 2, mo);
  }
  temp = 0;
  for (int i = 1; i <= n; i++) {
    temp += c(m, i - 1, mo);
    ans[i] = temp;
  }
  for (int i = n; i >= 1; i--) printf("%I64d ", quickpow(a, ans[i], mod));
}
