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;
inline int mul(int x, int y, int m) { return (long long)x * (long long)y % m; }
inline int add(int x, int y, int m) { return (x + y) % m; }
inline int qpow(int a, int p, int m) {
  int ans = 1, x = a % m;
  while (p) {
    if (p & 1) ans = mul(ans, x, m);
    x = mul(x, x, m);
    p >>= 1;
  }
  return ans;
}
int N, M, A, Q, SM;
inline int phi(int a) {
  int x, ans = a;
  for (x = 2; (long long)x * x <= a; x++)
    if (!(a % x)) {
      while (!(a % x)) a /= x;
      ans = ans / x * (x - 1);
    }
  if (a) ans = ans / a * (a - 1);
  return ans;
}
inline int stage(int a, int m) {
  int x, p = phi(m);
  for (x = 2; (long long)x * x <= p; x++)
    if (!(p % x)) {
      while (!(p % x)) p /= x;
      if (qpow(a, x, m) == 1) return x;
    }
  if (p && qpow(a, p, m) == 1) return p;
  return -1;
}
const int MAXM = 1000200;
int C[MAXM], S[MAXM];
int main() {
  int i;
  scanf("%d%d%d%d", &N, &M, &A, &Q);
  SM = stage(A, Q);
  C[0] = S[0] = 1;
  for (i = 1; i < N; i++)
    C[i] = mul(mul(C[i - 1], M - i + 1, SM), qpow(i, SM - 2, SM), SM),
    S[i] = add(S[i - 1], C[i], SM);
  for (i = N - 1; i >= 0; i--) printf("%d ", qpow(A, S[i], Q));
}
