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 INF = 0x3fffffff;
const int SINF = 0x7fffffff;
const long long LINF = 0x3fffffffffffffff;
const long long SLINF = 0x7fffffffffffffff;
const int MAXN = 1000207;
int n, m, a, q, MOD;
long long C[MAXN], pC[MAXN];
long long pp[MAXN];
int ans[MAXN];
void init();
void input();
void work();
long long qpow(long long a, long long b) {
  long long base = a % MOD, ans = 1;
  while (b) {
    if (b & 1) (ans *= base) %= MOD;
    (base *= base) %= MOD;
    b >>= 1;
  }
  return ans;
}
int main() {
  init();
  input();
  work();
}
void init() { ios::sync_with_stdio(false); }
void input() { scanf("%d%d%d%d", &n, &m, &a, &q); }
void work() {
  long long tmp = a;
  pp[0] = 1;
  for (int i = 1;; ++i) {
    pp[i] = tmp;
    if (tmp == 1) {
      MOD = i;
      break;
    }
    (tmp *= a) %= q;
  }
  C[0] = 1;
  for (int i = 1; i <= m; ++i)
    C[i] = ((C[i - 1] * (m - i + 1) % MOD) * qpow(i, MOD - 2) % MOD);
  pC[m] = 1;
  for (int i = m - 1; i >= 0; --i) pC[i] = (pC[i + 1] + C[i]) % MOD;
  int bs = 1;
  for (int i = 0; i < m; ++i) (bs <<= 1) %= MOD;
  for (int i = 1; i <= n; ++i) {
    ans[i] = ((bs - pC[n - i + 1]) % MOD + MOD) % MOD;
  }
  for (int i = 1; i <= n; ++i) printf("%I64d ", pp[ans[i]]);
  putchar('\n');
}
