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;
const int N = 1000005;
const int inf = 1e9 + 7;
long long fac[N], M, inv[N];
inline long long po(long long x, long long y, long long z) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % z;
    x = x * x % z;
    y >>= 1;
  }
  return res;
}
long long ans[N], M2;
inline long long C(int x, int y) {
  return fac[x] * inv[y] % M2 * inv[x - y] % M2;
}
int main() {
  int n, m, a;
  scanf("%d%d%d%lld", &n, &m, &a, &M);
  long long res = 1;
  for (int i = (1); i <= (2000000); i++) {
    res = res * (long long)a % M;
    if (res == 1) {
      M2 = i;
      break;
    }
  }
  fac[0] = fac[1] = inv[0] = inv[1] = 1;
  for (int i = (2); i <= (M2); i++) {
    fac[i] = fac[i - 1] * (long long)i % M2;
    inv[i] = po(fac[i], M2 - 2, M2);
  }
  ans[0] = 1;
  for (int i = (1); i <= (N - 1); i++) {
    ans[i] = ans[i - 1];
    ans[i] += C(m, i);
  }
  for (int i = (n - 1); i >= (0); i--) printf("%lld ", po(a, ans[i], M));
  puts("");
  return 0;
}
