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 = 1234567;
int dp[N], fac[N];
int mod, phi = -1;
long long pow(int a, int b, int MOD) {
  long long x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > MOD) x %= MOD;
    }
    y = (y * y);
    if (y > MOD) y %= MOD;
    b /= 2;
  }
  return x;
}
long long C(int m, int k) {
  long long den = (1LL * fac[k] * fac[m - k]) % phi;
  return (pow(den, phi - 2, phi) * fac[m]) % phi;
}
int main() {
  int n, m, a, i;
  long long x = 1;
  cin >> n >> m >> a >> mod;
  m++;
  fac[0] = 1;
  for (i = 1; i < N; i++) {
    x = (x * a) % mod;
    if (x == 1 && phi == -1) {
      phi = i;
      break;
    }
  }
  dp[0] = 1 % phi;
  dp[1] = m % phi;
  for (i = 1; i < N; i++) {
    fac[i] = (1LL * fac[i - 1] * i) % phi;
  }
  for (i = 2; i < n; i++) {
    if (i >= m)
      dp[i] = pow(2, m - 1, phi);
    else {
      dp[i] = (C(m, i) + dp[i - 2]) % phi;
    }
  }
  for (i = n - 1; i >= 0; i--) printf("%d ", pow(a, dp[i], mod));
}
