This is the easy version of the problem. The only difference between easy and hard versions is the constraint of m. You can make hacks only if both versions are solved.

Chiori loves dolls and now she is going to decorate her bedroom!

<image>

As a doll collector, Chiori has got n dolls. The i-th doll has a non-negative integer value a_i (a_i < 2^m, m is given). Chiori wants to pick some (maybe zero) dolls for the decoration, so there are 2^n different picking ways.

Let x be the bitwise-xor-sum of values of dolls Chiori picks (in case Chiori picks no dolls x = 0). The value of this picking way is equal to the number of 1-bits in the binary representation of x. More formally, it is also equal to the number of indices 0 ≤ i < m, such that \left⌊ (x)/(2^i) \right⌋ is odd.

Tell her the number of picking ways with value i for each integer i from 0 to m. Due to the answers can be very huge, print them by modulo 998 244 353.

Input

The first line contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5, 0 ≤ m ≤ 35) — the number of dolls and the maximum value of the picking way.

The second line contains n integers a_1, a_2, …, a_n (0 ≤ a_i < 2^m) — the values of dolls.

Output

Print m+1 integers p_0, p_1, …, p_m — p_i is equal to the number of picking ways with value i by modulo 998 244 353.

Examples

Input


4 4
3 5 8 14


Output


2 2 6 6 0 

Input


6 7
11 45 14 9 19 81


Output


1 2 11 20 15 10 5 0 
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("avx2")
using namespace std;
template <typename T>
void maxtt(T &t1, T t2) {
  t1 = max(t1, t2);
}
template <typename T>
void mintt(T &t1, T t2) {
  t1 = min(t1, t2);
}
bool debug = 0;
int n, m, k;
string direc = "RDLU";
const long long MOD2 = (long long)998244353 * (long long)998244353;
long long ln, lk, lm;
void etp(bool f = 0) {
  puts(f ? "YES" : "NO");
  exit(0);
}
void addmod(int &x, int y, int mod = 998244353) {
  x += y;
  if (x >= mod) x -= mod;
  if (x < 0) x += mod;
  assert(x >= 0 && x < mod);
}
void et(int x = -1) {
  printf("%d\n", x);
  exit(0);
}
long long fastPow(long long x, long long y, int mod = 998244353) {
  long long ans = 1;
  while (y > 0) {
    if (y & 1) ans = (x * ans) % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
long long gcd1(long long x, long long y) { return y ? gcd1(y, x % y) : x; }
long long a[200135];
struct lsp {
  long long a[60] = {0};
  const int maxBit = 54;
  bool insert(long long x) {
    for (int i = maxBit; ~i; i--)
      if (x & (1LL << i)) {
        if (a[i] != 0)
          x ^= a[i];
        else {
          for (int(j) = 0; (j) < (int)(i); (j)++)
            if (x & (1LL << j)) x ^= a[j];
          for (int j = i + 1; j <= maxBit; j++)
            if (a[j] & (1LL << i)) a[j] ^= x;
          a[i] = x;
          return 1;
        }
      }
    return 0;
  }
  lsp getOrthogonal(int m) {
    lsp res;
    vector<int> vp;
    for (int j = m - 1; j >= 0; j--)
      if (!a[j]) {
        vp.push_back(j);
        res.a[j] |= 1LL << j;
      }
    for (int j = m - 1; j >= 0; j--)
      if (a[j]) {
        int cc = 0;
        for (int z = m - 1; z >= 0; z--)
          if (!a[z]) {
            long long w = (a[j] >> z) & 1;
            res.a[vp[cc]] |= w << j;
            cc++;
          }
      }
    return res;
  }
} sp;
int p[66], q[66];
void ppt() {
  for (int i = 0; i <= m; i++) {
    printf("%lld ", (long long)p[i] * fastPow(2, n - k) % 998244353);
  }
  exit(0);
}
vector<long long> bs;
inline void dfs(int *p, int k, int i, long long x) {
  if (i == k) {
    p[__builtin_popcountll(x)]++;
    return;
  }
  dfs(p, k, i + 1, x);
  dfs(p, k, i + 1, x ^ bs[i]);
}
void calsm(lsp sp, int *p, int k) {
  bs.clear();
  for (int(j) = 0; (j) < (int)(sp.maxBit); (j)++)
    if (sp.a[j]) bs.push_back(sp.a[j]);
  assert(bs.size() == k);
  dfs(p, k, 0, 0);
}
void calbg() {
  vector<vector<int>> C(60, vector<int>(60, 0));
  C[0][0] = 1;
  for (int i = 1; i <= 55; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
    }
  }
  vector<vector<int>> w(m + 1, vector<int>(m + 1, 0));
  for (int c = 0; c <= m; c++) {
    for (int d = 0; d <= m; d++) {
      for (int j = 0; j <= c; j++) {
        int val = (long long)C[d][j] * C[m - d][c - j] % 998244353;
        if (j % 2 == 0)
          addmod(w[c][d], val);
        else
          addmod(w[c][d], 998244353 - val);
      }
    }
  }
  lsp B = sp.getOrthogonal(m);
  calsm(B, q, m - k);
  for (int(c) = 0; (c) < (int)(m + 1); (c)++) {
    for (int d = 0; d <= m; d++) {
      addmod(p[c], (long long)q[d] * w[c][d] % 998244353);
    }
  }
  int tmp = fastPow(2, m - k);
  tmp = fastPow(tmp, 998244353 - 2);
  for (int(c) = 0; (c) < (int)(m + 1); (c)++)
    p[c] = (long long)p[c] * tmp % 998244353;
  ppt();
}
void fmain(int tid) {
  scanf("%d%d", &n, &m);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%lld", a + i);
  for (int(i) = 1; (i) <= (int)(n); (i)++) k += sp.insert(a[i]);
  if (k <= m / 2) {
    calsm(sp, p, k);
    ppt();
  }
  calbg();
}
int main() {
  int t = 1;
  for (int(i) = 1; (i) <= (int)(t); (i)++) {
    fmain(i);
  }
  return 0;
}
