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 long long inf = (long long)1e18;
const long long mod = (long long)1e9 + 7;
const double pi = acos(-1.0);
const double eps = (double)1e-9;
const int dx[] = {0, 0, 1, 0, -1};
const int dy[] = {0, 1, 0, -1, 0};
const int N = 2000500;
int n, m, a, Q, phi, f[N], rv[N], cur, last, e[N];
int bp(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res = 1LL * res * x % Q;
    x = 1LL * x * x % Q;
    y >>= 1;
  }
  return res;
}
int cnk(int n, int k) {
  if (n < k) return 0;
  return 1LL * f[n] * rv[k] % Q * rv[n - k] % Q;
}
inline int gcd(int a, int b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
int main() {
  cin >> n >> m >> a >> Q;
  for (int i = 1; i < Q; ++i)
    if (bp(a, i) == 1) {
      phi = i;
      break;
    }
  swap(phi, Q);
  f[0] = 1;
  rv[0] = 1;
  for (int i = 1; i <= 2e6; ++i) {
    f[i] = 1LL * f[i - 1] * i % Q;
    rv[i] = bp(f[i], Q - 2);
  }
  for (int i = 1; i <= min(n, m + 1); ++i) {
    cur = (cur + cnk(m, i - 1)) % Q;
    last = i;
  }
  last--;
  e[1] = cur;
  for (int i = 2; i + m <= n; ++i) e[i] = cur;
  for (int i = max(2, n - m + 1); i <= n; ++i) {
    cur = (cur + Q - cnk(m, last--)) % Q;
    e[i] = cur;
  }
  swap(phi, Q);
  for (int i = 1; i <= n; ++i) printf("%d ", bp(a, e[i]));
  return 0;
}
