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;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
string direc = "RDLU";
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 1000000007) {
  assert(y >= 0);
  x += y;
  if (x >= mod) x -= mod;
  assert(x >= 0 && x < mod);
}
void et() {
  puts("-1");
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 1000000007) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
int a, Q, P = 1;
int fac[1000151], inv[1000151];
void initfac(int n) {
  fac[0] = inv[0] = 1;
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    fac[i] = (long long)fac[i - 1] * i % P;
    inv[i] = fastPow(fac[i], P - 2, P);
  }
}
int C(int n, int m) {
  if (n < m) return 0;
  return (long long)fac[n] * inv[m] % P * inv[n - m] % P;
}
int ans[1000151], ta[1000151];
void fmain(int tid) {
  scanf("%d%d%d%d", &n, &m, &a, &Q);
  ta[0] = 1;
  for (ta[P] = a % Q; ta[P] != 1;) {
    P++;
    ta[P] = (long long)ta[P - 1] * a % Q;
  }
  initfac(m);
  for (int i = n, j = 0; i; i--, j++) {
    ans[i] = (ans[i + 1] + C(m, j)) % P;
  }
  for (int(i) = 1; (i) <= (int)(n); (i)++) {
    printf("%d ", ta[ans[i]]);
  }
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
