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 mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1.0);
const double EPS = 1e-10;
inline int readint() {
  int sum = 0;
  char c = getchar();
  bool f = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return f ? -sum : sum;
}
inline long long readLL() {
  long long sum = 0;
  char c = getchar();
  bool f = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    sum = sum * 10 + c - '0';
    c = getchar();
  }
  return f ? -sum : sum;
}
long long qmod(long long a, long long b, long long c) {
  long long ret = 1;
  for (; b; b >>= 1) {
    if (b & 1) ret = ret * a % c;
    a = a * a % c;
  }
  return ret;
}
const int maxn = 1000125;
long long inv[maxn], ami[maxn], x[maxn];
void Inverseinit(int MAXNUM, long long mod) {
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= MAXNUM; i++)
    inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
int main() {
  int n = readint(), m = readint(), a = readint(), q = readint();
  ami[0] = 1;
  int phi;
  for (int i = 1;; i++) {
    ami[i] = ami[i - 1] * a % q;
    if (ami[i] == 1) {
      phi = i;
      break;
    }
  }
  Inverseinit(n, phi);
  x[0] = 1;
  m++;
  for (int i = 1; i < n; i++)
    x[i] = x[i - 1] * (m + 1 - i) % phi * inv[i] % phi,
    x[i] = (x[i] + phi) % phi;
  for (int i = 2; i < n; i++) x[i] = (x[i] + x[i - 2]) % phi;
  for (int i = n - 1; i >= 0; i--) printf("%lld ", ami[x[i]]);
  printf("\n");
  return 0;
}
