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>
using namespace std;
using uint = unsigned int;
using ll = long long;
using ull = unsigned long long;
template <uint MD>
struct ModInt {
  using M = ModInt;
  static constexpr uint get_mod() { return MD; }
  const static M G;
  uint v;
  ModInt(ll _v = 0) { set_v(uint(_v % MD + MD)); }
  M& set_v(uint _v) {
    v = (_v < MD) ? _v : _v - MD;
    return *this;
  }
  explicit operator bool() const { return v != 0; }
  M operator-() const { return M() - *this; }
  M operator+(const M& r) const { return M().set_v(v + r.v); }
  M operator-(const M& r) const { return M().set_v(v + MD - r.v); }
  M operator*(const M& r) const { return M().set_v(uint(ull(v) * r.v % MD)); }
  M operator/(const M& r) const { return *this * r.inv(); }
  M& operator+=(const M& r) { return *this = *this + r; }
  M& operator-=(const M& r) { return *this = *this - r; }
  M& operator*=(const M& r) { return *this = *this * r; }
  M& operator/=(const M& r) { return *this = *this / r; }
  bool operator==(const M& r) const { return v == r.v; }
  M pow(ll n) const {
    M x = *this, r = 1;
    while (n) {
      if (n & 1) r *= x;
      x *= x;
      n >>= 1;
    }
    return r;
  }
  M inv() const { return pow(MD - 2); }
  friend ostream& operator<<(ostream& os, const M& r) { return os << r.v; }
};
using Mint = ModInt<998244353>;
int cnt[17][13];
Mint poss[17][65536];
Mint ans[17][65536];
Mint fac[65540];
Mint ifac[65540];
Mint possin[17][17][2][16];
void transform(Mint* from, Mint* to) {
  if (to - from == 1) return;
  Mint* mid = from + (to - from) / 2;
  transform(from, mid);
  transform(mid, to);
  for (int i = 0; i < mid - from; i++) {
    Mint a = *(from + i);
    Mint b = *(mid + i);
    *(from + i) = a + b;
    *(mid + i) = a - b;
  }
}
Mint C(int n, int k) { return fac[n] * ifac[n - k] * ifac[k]; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  fac[0] = 1;
  ifac[0] = 1;
  for (int i = 1; i <= 65536; i++) {
    fac[i] = fac[i - 1] * i;
    ifac[i] = ifac[i - 1] * Mint(i).inv();
  }
  int n, k, oc;
  cin >> n >> k >> oc;
  int c = max(oc, 4);
  int allx = 0;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    allx ^= a;
    int low = a & 15;
    int high = (a >> 4);
    int highbit = (high == 0 ? c - 4 : __builtin_ctz(high));
    cnt[low][highbit]++;
  }
  ans[0][0] = 1;
  for (int j = 0; j <= 16; j++) {
    transform(ans[j], ans[j] + (1 << (c - 1)));
  }
  Mint invc = Mint(1 << (c - 1)).inv();
  for (int low = 0; low < 16; low++) {
    for (int highbit = 0; highbit <= c - 4; highbit++)
      if (cnt[low][highbit]) {
        memset(possin, 0, sizeof(possin));
        memset(poss, 0, sizeof(poss));
        possin[0][0][0][0] = 1;
        for (int guy = 1; guy <= min(k, cnt[low][highbit]); guy++) {
          for (int moves = 1; moves <= k + 1 - guy; moves++) {
            for (int prevm = guy - 1; prevm + moves <= k; prevm++) {
              for (int prevxi = 0; prevxi <= 1; prevxi++) {
                for (int prevxl = 0; prevxl < 16; prevxl++) {
                  if (possin[guy - 1][prevm][prevxi][prevxl].v) {
                    int xi = moves > low;
                    int xl = ((((1 << 20) + low) - moves) & 15) ^ low;
                    possin[guy][prevm + moves][prevxi ^ xi][prevxl ^ xl] +=
                        possin[guy - 1][prevm][prevxi][prevxl] *
                        C(prevm + moves, moves);
                  }
                }
              }
            }
          }
        }
        int xix = ((1 << (highbit + 1)) - 1) << 4;
        for (int guy = 1; guy <= min(k, cnt[low][highbit]); guy++) {
          for (int moves = guy; moves <= k; moves++) {
            for (int prevxi = 0; prevxi <= 1; prevxi++) {
              if (prevxi == 1 && moves <= low) continue;
              for (int prevxl = 0; prevxl < 16; prevxl++) {
                poss[moves][(prevxl ^ (prevxi ? xix : 0)) >> 1] +=
                    possin[guy][moves][prevxi][prevxl] *
                    C(cnt[low][highbit], guy);
              }
            }
          }
        }
        for (int moves = 1; moves <= k; moves++) {
          transform(poss[moves], poss[moves] + (1 << (c - 1)));
        }
        for (int tot = k; tot >= 1; tot--) {
          for (int moves = 1; moves <= tot; moves++) {
            for (int i = 0; i < (1 << (c - 1)); i++) {
              ans[tot][i] +=
                  ans[tot - moves][i] * poss[moves][i] * C(tot, moves);
            }
          }
        }
      }
  }
  Mint denom = 1;
  Mint invn = Mint(n).inv();
  for (int i = 0; i < k; i++) {
    denom *= invn;
  }
  transform(ans[k], ans[k] + (1 << (c - 1)));
  for (int i = 0; i < (1 << oc); i++) {
    if ((i & 1) == ((allx & 1) ^ (k & 1))) {
      cout << ans[k][(i >> 1) ^ (allx >> 1)] * denom * invc << " ";
    } else
      cout << 0 << " ";
  }
  cout << endl;
  return 0;
}