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;
int n, m, a, Q, Z;
long long fac[1000000 + 130], inv[1000000 + 130], c[1000000 + 130];
void factorial() {
  int top = 1e6 + 123;
  fac[0] = 1;
  for (int i = 1; i <= top; i++) fac[i] = fac[i - 1] * i % Z;
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= top; i++) inv[i] = inv[Z % i] * (Z - Z / i) % Z;
  for (int i = 2; i <= top; i++) inv[i] = inv[i] * inv[i - 1] % Z;
}
long long C(int n, int m) { return fac[n] * inv[m] % Z * inv[n - m] % Z; }
long long quick(long long a, long long b, long long M) {
  long long c = 1;
  while (b) {
    if (b & 1) c = c * a % M;
    a = a * a % M;
    b >>= 1;
  }
  return c;
}
long long ans[1000000 + 130];
int main() {
  scanf("%d%d%d%d", &n, &m, &a, &Q);
  Z = 1;
  while (quick(a, Z, Q) != 1) Z++;
  factorial();
  for (int i = 1; i <= m + 1; i++) {
    c[i] = C(m, i - 1);
    c[i] = (c[i] + c[i - 1]) % Z;
  }
  for (int i = n; i; i--) {
    if (n - i + 1 <= m + 1)
      ans[i] = c[n - i + 1];
    else
      ans[i] = ans[i + 1];
  }
  for (int i = 1; i < n; i++) printf("%lld ", quick(a, ans[i], Q));
  printf("%lld\n", quick(a, ans[n], Q));
  return 0;
}
