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 inv[2100000 + 1];
long long kai[2100000 + 1];
long long invkai[2100000 + 1];
void invinit() {
  inv[1] = 1;
  for (int i = 2; i <= 2100000; i++) {
    inv[i] = mod - (mod / i) * inv[mod % i] % mod;
  }
  kai[0] = invkai[0] = 1;
  for (int i = 1; i <= 2100000; i++) {
    kai[i] = kai[i - 1] * i % mod;
    invkai[i] = invkai[i - 1] * inv[i] % mod;
  }
}
long long com(long long a, long long b) {
  if (b < 0 || a < b) return 0;
  return (invkai[b] * invkai[a - b]) % mod * kai[a] % mod;
}
long long po(long long a, long long b, long long m) {
  if (b == 0) return 1;
  long long t = po(a, b / 2, m);
  t = t * t % m;
  if (b & 1) t = t * a % m;
  return t;
}
long long ans[2100000];
int main() {
  int num, kai, st, m;
  scanf("%d%d%d%d", &num, &kai, &st, &m);
  int c = 0;
  long long now = 1;
  for (;;) {
    now = now * st % m;
    c++;
    if (now == 1) {
      mod = c;
      break;
    }
  }
  invinit();
  ans[0] = 1;
  for (int i = 1; i < num; i++) ans[i] = (ans[i - 1] + com(kai, i)) % mod;
  for (int i = num - 1; i >= 0; i--) {
    printf("%I64d ", po(st, ans[i], m));
  }
  printf("\n");
}
