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 long long INF = 0x3f3f3f3f3f3f3f3f;
const long long inf = 0xc0c0c0c0c0c0c0c0;
const double pi = acos(-1);
long long N, M, a, Q, mod;
const int MAXN = 1e6 + 200;
long long ans[MAXN];
long long tool[MAXN];
long long fact[MAXN];
long long qpow(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * b % mod;
    p >>= 1;
    b = b * b % mod;
  }
  return ans;
}
void init() {
  fact[0] = 1;
  for (long long i = 1; i <= M; i++) fact[i] = fact[i - 1] * i % mod;
}
long long C(long long b, long long s) {
  if (b < s) return 0;
  return fact[b] * qpow(fact[b - s] * fact[s] % mod, mod - 2) % mod;
}
int main() {
  scanf("%lld%lld%lld%lld", &N, &M, &a, &Q);
  mod = 1;
  long long cpy = a;
  tool[0] = 1;
  while (cpy != 1) {
    tool[mod] = cpy;
    cpy = cpy * a % Q;
    mod++;
  }
  init();
  ans[0] = 0;
  for (int i = 1; i <= N; i++) ans[i] = (ans[i - 1] + C(M, i - 1)) % mod;
  for (int i = N; i; i--) printf("%lld ", tool[ans[i]]);
  return 0;
}
