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;
template <typename T>
inline T qmin(const T &a, const T &b) {
  return a < b ? a : b;
}
template <typename T>
inline T qmax(const T &a, const T &b) {
  return a > b ? a : b;
}
template <typename T>
inline void getmin(T &a, const T &b) {
  if (a > b) a = b;
}
template <typename T>
inline void getmax(T &a, const T &b) {
  if (a < b) a = b;
}
inline void fileio(string s) {
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
const int inf = (int)1e9 + 7;
const long long linf = (long long)1e17 + 7;
const int N = 2e6 + 7;
int mod, phi;
long long fac[N], inv[N], sum[N];
inline long long qpow(long long x, long long y, const long long &mod) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
int n, m, a;
int main() {
  scanf("%d%d%d%d", &n, &m, &a, &mod);
  phi = 1;
  long long cur = a;
  while (cur != 1) {
    cur = cur * a % mod;
    ++phi;
  }
  fac[0] = inv[0] = 1;
  for (register int i = (1); i <= (m); ++i) {
    fac[i] = fac[i - 1] * i % phi;
    inv[i] = qpow(fac[i], phi - 2, phi);
  }
  for (register int i = 0; i < (n); ++i) {
    sum[i] = (sum[i - 1] + fac[m] * inv[m - i] % phi * inv[i] % phi) % phi;
  }
  for (register int i = (1); i <= (n); ++i)
    printf("%lld ", qpow(a, sum[n - i], mod));
  puts("");
  return 0;
}
