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 = 1000000007;
long long POW(long long a, long long b, long long MMM = MOD) {
  long long ret = 1;
  for (; b; b >>= 1, a = (a * a) % MMM)
    if (b & 1) ret = (ret * a) % MMM;
  return ret;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) {
  if (a == 0 || b == 0) return a + b;
  return a * (b / gcd(a, b));
}
int dx[] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[] = {1, 0, -1, 0, 1, -1, 1, -1};
int n, m;
long long a, p;
long long fac[1000200];
long long inv[1000200];
long long nCr(int n, int r) {
  if (r > n) return 0;
  long long c = fac[n];
  c = (c * inv[r]) % p;
  c = (c * inv[n - r]) % p;
  return c;
}
long long phi(long long a) {
  long long c = 1;
  for (long long i = 1;; i++) {
    c = (c * a) % MOD;
    if (c == 1) return i;
  }
}
long long ans[1000200];
int main() {
  scanf("%d%d%lld%lld", &n, &m, &a, &MOD);
  p = phi(a);
  fac[0] = inv[0] = 1;
  for (int(i) = (1); (i) <= (p - 1); (i) += (1)) fac[i] = (fac[i - 1] * i) % p;
  inv[p - 1] = POW(fac[p - 1], p - 2, p);
  for (int(i) = (p - 2); (i) >= (1); (i) -= (1))
    inv[i] = (inv[i + 1] * (i + 1)) % p;
  for (int(i) = (n - 1); (i) >= (0); (i) -= (1)) {
    ans[i] = (ans[i + 1] + nCr(m, n - 1 - i)) % p;
  }
  for (int(i) = (0); (i) <= (n - 1); (i) += (1))
    printf("%lld ", POW(a, ans[i]));
}
