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>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
const int N = (1 << 16) + 5, mod = 998244353;
template <typename _Tp1, typename _Tp2>
inline void add(_Tp1 &a, _Tp2 b) {
  (a += b) >= mod && (a -= mod);
}
template <typename _Tp1, typename _Tp2>
inline void sub(_Tp1 &a, _Tp2 b) {
  (a -= b) < 0 && (a += mod);
}
long long ksm(long long a, long long b = mod - 2) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod, b >>= 1;
  }
  return res;
}
long long fac[18], inv[18], ifac[18];
int a[18], b[18], e[N][17], cnt[1 << 16], f[N][17];
std::array<int, 16> d[N];
int main() {
  for (int i = 1; i < 1 << 16; ++i) cnt[i] = cnt[i >> 1] + (i & 1);
  fac[0] = fac[1] = inv[0] = inv[1] = ifac[0] = ifac[1] = 1;
  for (int i = 2; i < 18; ++i)
    fac[i] = fac[i - 1] * i % mod,
    inv[i] = inv[mod % i] * (mod - mod / i) % mod,
    ifac[i] = ifac[i - 1] * inv[i] % mod;
  int n, k, c, s = 0;
  read(n, k, c);
  for (int st = 0; st < 1 << k; ++st) {
    a[0] = 1;
    for (int j = 0; j < k; ++j)
      a[j + 1] = st >> j & 1 ? mod - ifac[j + 1] : ifac[j + 1];
    b[0] = a[1];
    for (int i = 1; i < k; ++i) {
      unsigned long long v = 0;
      for (int j = 0; j < i; ++j) v += 1ULL * b[j] * a[i - j];
      b[i] = (1LL * (i + 1) * a[i + 1] + mod - v % mod) % mod;
    }
    for (int i = k - 1; i >= 0; --i) b[i + 1] = 1ULL * b[i] * inv[i + 1] % mod;
    b[0] = 0;
    for (int i = 0; i <= k; ++i) e[st][i] = b[i];
  }
  for (int i = 1, x; i <= n; ++i) {
    read(x), s ^= x;
    for (int j = 0; j < k; ++j) d[i][j] = x ^ (x - j - 1);
  }
  std::sort(d + 1, d + n + 1);
  for (int i = 1; i <= n; ++i) {
    int _ = i;
    while (_ < n && d[_ + 1] == d[i]) ++_;
    for (int j = 0; j < 1 << c; ++j) {
      int st = 0;
      for (int p = 0; p < k; ++p) st |= (cnt[d[i][p] & j] & 1) << p;
      for (int p = 0; p <= k; ++p)
        add(f[j][p], 1LL * (_ - i + 1) * e[st][p] % mod);
    }
    i = _;
  }
  for (int st = 0; st < 1 << c; ++st) {
    for (int i = 0; i < k; ++i) a[i] = 1ULL * (i + 1) * f[st][i + 1] % mod;
    b[0] = 1;
    for (int i = 0; i < k; ++i) {
      unsigned long long v = 0;
      for (int j = 0; j <= i; ++j) v += 1ULL * b[j] * a[i - j];
      b[i + 1] = 1ULL * v % mod * inv[i + 1] % mod;
    }
    for (int i = 0; i <= k; ++i) f[st][i] = b[i];
  }
  for (int len = 1; len < 1 << c; len <<= 1)
    for (int i = 0; i < 1 << c; i += len << 1)
      for (int j = 0; j < len; ++j)
        for (int d = 0; d <= k; ++d) {
          int tmp = f[i + j + len][d];
          f[i + j + len][d] = f[i + j][d] - tmp < 0 ? f[i + j][d] - tmp + mod
                                                    : f[i + j][d] - tmp,
                    add(f[i + j][d], tmp);
        }
  long long inv = (mod - ((mod - 1) >> c)) * ksm(ksm(n), k) % mod;
  for (int i = 0; i < 1 << c; ++i)
    printf("%lld%c", fac[k] * f[s ^ i][k] % mod * inv % mod,
           " \n"[i + 1 == (1 << c)]);
  return 0;
}