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;
long long fac[1000150], inv[1000150];
long long bigmod(long long b, int i, long long mod) {
  long long r = 1;
  for (int j = (29); j >= (0); j--) {
    r = r * r % mod;
    if ((i >> j) & 1) r = r * b % mod;
  }
  return r;
}
long long nCr(int n, int r) { return fac[n] * inv[r] % mod * inv[n - r] % mod; }
long long output[1000150];
int main() {
  int n, m, a, q;
  scanf("%d %d %d %d", &n, &m, &a, &q);
  long long t = a;
  mod = 1;
  while (t != 1) {
    t = t * a % q;
    ++mod;
  }
  fac[0] = inv[0] = 1;
  for (int i = (1); i <= (m); i++) {
    fac[i] = fac[i - 1] * i % mod;
    inv[i] = bigmod(fac[i], mod - 2, mod);
  }
  long long sum = 0;
  for (int i = (1); i <= (n); i++) {
    if (i <= m + 1) {
      sum = (sum + nCr(m, i - 1)) % mod;
    }
    output[n - i + 1] = bigmod(a, sum, q);
  }
  for (int i = (1); i <= (n); i++) {
    printf("%lld%c", output[i], (i == n) ? '\n' : ' ');
  }
  return 0;
}
