Problem: You are given a positive integer k and an array a_1, a_2, …, a_n of non-negative distinct integers not smaller than k and not greater than 2^c-1.

In each of the next k seconds, one element is chosen randomly equiprobably out of all n elements and decreased by 1.

For each integer x, 0 ≤ x ≤ 2^c - 1, you need to find the probability that in the end the bitwise XOR of all elements of the array is equal to x. 

Each of these values can be represented as an irreducible fraction p/q, and you need to find the value of p ⋅ q^{-1} modulo 998 244 353. 

Input

The first line of input contains three integers n, k, c (1 ≤ n ≤ (2^c - k), 1 ≤ k ≤ 16, 1 ≤ c ≤ 16).

The second line contains n distinct integers a_1, a_2, …, a_n (k ≤ a_i ≤ 2^c-1).

Output

Print 2^c integers: the probability that the bitwise XOR is equal to x in the end for x in \{0, 1, …, 2^c-1\} modulo 998 244 353.

Example

Input


4 1 3
1 2 3 4


Output


0 0 0 748683265 0 499122177 0 748683265 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimization("unroll-all-loops")
using namespace std;
const int kMod = 998244353;
struct barrett {
  using uint = unsigned int;
  using ull = unsigned long long;
  uint m;
  ull im;
  barrett(uint m) : m(m), im((ull)(-1) / m + 1) {}
  uint mul(uint a, uint b) const {
    ull z = a;
    z *= b;
    ull x = (ull)(((unsigned __int128)(z)*im) >> 64);
    uint v = (uint)(z - x * m);
    if (m <= v) v += m;
    return v;
  }
};
barrett bt(kMod);
int inv(int b) {
  int e = kMod - 2;
  int r = 1;
  while (e) {
    if (e % 2) r = bt.mul(r, b);
    b = bt.mul(b, b);
    e /= 2;
  }
  return r;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n, k, c;
  cin >> n >> k >> c;
  vector<int> a(n), mx(n, 0);
  int all = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    for (int j = 0; j <= k; ++j) {
      mx[i] = max(mx[i], (a[i] ^ (a[i] - j)));
    }
    all ^= a[i];
  }
  vector<int> order(n);
  iota(order.begin(), order.end(), 0);
  sort(order.begin(), order.end(), [&](int a, int b) { return mx[a] < mx[b]; });
  vector<vector<int>> comb(2 * k + 1, vector<int>(2 * k + 1, 0));
  for (int i = 0; i <= 2 * k; ++i) {
    comb[i][0] = comb[i][i] = 1;
    for (int j = 1; j < i; ++j) {
      comb[i][j] = (comb[i - 1][j - 1] + comb[i - 1][j]) % kMod;
    }
  }
  vector<vector<long long>> dp(k + 1, vector<long long>(1 << c, 0));
  dp[0][0] = 1;
  int lim = 1;
  for (auto it : order) {
    for (int j = k; j >= 0; --j) {
      for (int i = 1; i + j <= k; ++i) {
        int val = (a[it] ^ (a[it] - i)) / 2;
        for (int x = 0; x < lim; ++x)
          dp[i + j][x ^ val] += bt.mul(comb[i + j][j], dp[j][x]);
      }
    }
    for (int j = 0; j <= k; ++j) {
      int val = (a[it] ^ (a[it] - j)) / 2;
      while (lim <= val) lim *= 2;
    }
    for (int j = 0; j <= k; ++j)
      for (int x = 0; x < lim; ++x) dp[j][x] %= kMod;
  }
  int sub = 1, invn = inv(n);
  for (int i = 0; i < k; ++i) sub = bt.mul(sub, invn);
  for (int i = 0; i < (1 << c); ++i) {
    int par = (k + all + i) % 2;
    if (par)
      cout << 0 << " ";
    else
      cout << bt.mul(dp[k][(i ^ all) / 2], sub) << " ";
  }
  cout << endl;
  return 0;
}