#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T read() {
  T f = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) f = c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  return f ? -x : x;
}
namespace run {
const int mod = 998244353, N = 1 << 16, inv2 = (mod + 1) / 2;
inline int add(int x, int y) { return x + y >= mod ? x - mod + y : x + y; }
inline int sub(int x, int y) { return x >= y ? x - y : x + mod - y; }
inline int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1LL * x * ret % mod;
    x = 1LL * x * x % mod, y >>= 1;
  }
  return ret;
}
int inv[17], fac[17], ifac[17];
inline void Getln(int *a, int *b, int len) {
  assert(a[0] == 1), b[0] = 0;
  for (int i = 0; i < len - 1; i++) {
    int sum = 1LL * a[i + 1] * (i + 1) % mod;
    for (int j = 0; j < i; j++)
      sum = sub(sum, 1LL * a[i - j] * (j + 1) % mod * b[j + 1] % mod);
    b[i + 1] = 1LL * sum * inv[i + 1] % mod;
  }
}
inline void Getexp(int *a, int *b, int len) {
  assert(a[0] == 0), b[0] = 1;
  for (int i = 0; i < len - 1; i++) {
    b[i + 1] = 0;
    for (int j = 0; j <= i; j++)
      b[i + 1] = (1LL * (j + 1) * a[j + 1] % mod * b[i - j] + b[i + 1]) % mod;
    b[i + 1] = 1LL * inv[i + 1] * b[i + 1] % mod;
  }
}
inline void fwt_xor(int *a, int inv, int limit) {
  for (int mid = 1; mid < limit; mid <<= 1)
    for (int i = 0; i < limit; i += (mid << 1))
      for (int j = 0; j < mid; j++) {
        int x = a[i + j], y = a[i + j + mid];
        if (inv == -1) x = 1LL * inv2 * x % mod, y = 1LL * inv2 * y % mod;
        a[i + j] = add(x, y), a[i + j + mid] = sub(x, y);
      }
}
int a[N], n, f[N], K, C, vis[N];
map<vector<int>, int> mp;
vector<int> st[209];
int count[N], tot, top, num[209], r[209], A[209];
int main() {
  n = read<int>(), K = read<int>(), C = read<int>();
  inv[0] = inv[1] = fac[0] = ifac[0] = ifac[1] = 1, count[0] = 0;
  for (int i = 1; i <= K; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int i = 2; i <= K; i++)
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod,
    ifac[i] = 1LL * inv[i] * ifac[i - 1] % mod;
  for (int i = 1; i < 1 << C; i++) count[i] = count[i >> 1] + (i & 1);
  int xorsum = 0;
  for (int i = 0; i < n; i++) xorsum ^= (a[i] = read<int>());
  for (int i = 0; i < n; i++) {
    vector<int> b;
    for (int j = 0; j <= K; j++) b.push_back(a[i] ^ (a[i] - j));
    mp[b]++;
  }
  for (map<vector<int>, int>::iterator i = mp.begin(); i != mp.end(); i++)
    st[++top] = i->first, num[top] = i->second;
  for (int s = 0; s < 1 << C; s++) {
    tot = 0;
    for (int i = 1; i <= top; i++) {
      int tmp = 0;
      for (int j = 0; j <= K; j++) tmp = (tmp << 1) | (count[st[i][j] & s] & 1);
      if (!vis[tmp])
        A[++tot] = tmp, r[tot] = num[i], vis[tmp] = tot;
      else
        r[vis[tmp]] += num[i];
    }
    static int c[17], d[17], ret[17];
    memset(ret, 0, sizeof ret);
    for (int i = 1; i <= tot; i++) {
      for (int j = 0; j <= K; j++)
        c[j] = (A[i] >> (K - j) & 1) ? mod - ifac[j] : ifac[j];
      Getln(c, d, K + 1);
      for (int j = 0; j <= K; j++) ret[j] = (1LL * d[j] * r[i] + ret[j]) % mod;
      vis[A[i]] = 0;
    }
    Getexp(ret, c, K + 1), f[s] = 1LL * c[K] * fac[K] % mod;
  }
  fwt_xor(f, -1, 1 << C);
  int p = qpow(qpow(n, K), mod - 2);
  for (int i = 0; i < 1 << C; i++)
    printf("%lld ", 1LL * f[i ^ xorsum] * p % mod);
  return 0;
}
}  // namespace run
int main() { return run::main(); }

### Prompt

In CPP, your task is to solve the following 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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T read() {
  T f = 0, x = 0;
  char c = getchar();
  while (!isdigit(c)) f = c == '-', c = getchar();
  while (isdigit(c)) x = x * 10 + c - 48, c = getchar();
  return f ? -x : x;
}
namespace run {
const int mod = 998244353, N = 1 << 16, inv2 = (mod + 1) / 2;
inline int add(int x, int y) { return x + y >= mod ? x - mod + y : x + y; }
inline int sub(int x, int y) { return x >= y ? x - y : x + mod - y; }
inline int qpow(int x, int y) {
  int ret = 1;
  while (y) {
    if (y & 1) ret = 1LL * x * ret % mod;
    x = 1LL * x * x % mod, y >>= 1;
  }
  return ret;
}
int inv[17], fac[17], ifac[17];
inline void Getln(int *a, int *b, int len) {
  assert(a[0] == 1), b[0] = 0;
  for (int i = 0; i < len - 1; i++) {
    int sum = 1LL * a[i + 1] * (i + 1) % mod;
    for (int j = 0; j < i; j++)
      sum = sub(sum, 1LL * a[i - j] * (j + 1) % mod * b[j + 1] % mod);
    b[i + 1] = 1LL * sum * inv[i + 1] % mod;
  }
}
inline void Getexp(int *a, int *b, int len) {
  assert(a[0] == 0), b[0] = 1;
  for (int i = 0; i < len - 1; i++) {
    b[i + 1] = 0;
    for (int j = 0; j <= i; j++)
      b[i + 1] = (1LL * (j + 1) * a[j + 1] % mod * b[i - j] + b[i + 1]) % mod;
    b[i + 1] = 1LL * inv[i + 1] * b[i + 1] % mod;
  }
}
inline void fwt_xor(int *a, int inv, int limit) {
  for (int mid = 1; mid < limit; mid <<= 1)
    for (int i = 0; i < limit; i += (mid << 1))
      for (int j = 0; j < mid; j++) {
        int x = a[i + j], y = a[i + j + mid];
        if (inv == -1) x = 1LL * inv2 * x % mod, y = 1LL * inv2 * y % mod;
        a[i + j] = add(x, y), a[i + j + mid] = sub(x, y);
      }
}
int a[N], n, f[N], K, C, vis[N];
map<vector<int>, int> mp;
vector<int> st[209];
int count[N], tot, top, num[209], r[209], A[209];
int main() {
  n = read<int>(), K = read<int>(), C = read<int>();
  inv[0] = inv[1] = fac[0] = ifac[0] = ifac[1] = 1, count[0] = 0;
  for (int i = 1; i <= K; i++) fac[i] = 1LL * fac[i - 1] * i % mod;
  for (int i = 2; i <= K; i++)
    inv[i] = 1LL * (mod - mod / i) * inv[mod % i] % mod,
    ifac[i] = 1LL * inv[i] * ifac[i - 1] % mod;
  for (int i = 1; i < 1 << C; i++) count[i] = count[i >> 1] + (i & 1);
  int xorsum = 0;
  for (int i = 0; i < n; i++) xorsum ^= (a[i] = read<int>());
  for (int i = 0; i < n; i++) {
    vector<int> b;
    for (int j = 0; j <= K; j++) b.push_back(a[i] ^ (a[i] - j));
    mp[b]++;
  }
  for (map<vector<int>, int>::iterator i = mp.begin(); i != mp.end(); i++)
    st[++top] = i->first, num[top] = i->second;
  for (int s = 0; s < 1 << C; s++) {
    tot = 0;
    for (int i = 1; i <= top; i++) {
      int tmp = 0;
      for (int j = 0; j <= K; j++) tmp = (tmp << 1) | (count[st[i][j] & s] & 1);
      if (!vis[tmp])
        A[++tot] = tmp, r[tot] = num[i], vis[tmp] = tot;
      else
        r[vis[tmp]] += num[i];
    }
    static int c[17], d[17], ret[17];
    memset(ret, 0, sizeof ret);
    for (int i = 1; i <= tot; i++) {
      for (int j = 0; j <= K; j++)
        c[j] = (A[i] >> (K - j) & 1) ? mod - ifac[j] : ifac[j];
      Getln(c, d, K + 1);
      for (int j = 0; j <= K; j++) ret[j] = (1LL * d[j] * r[i] + ret[j]) % mod;
      vis[A[i]] = 0;
    }
    Getexp(ret, c, K + 1), f[s] = 1LL * c[K] * fac[K] % mod;
  }
  fwt_xor(f, -1, 1 << C);
  int p = qpow(qpow(n, K), mod - 2);
  for (int i = 0; i < 1 << C; i++)
    printf("%lld ", 1LL * f[i ^ xorsum] * p % mod);
  return 0;
}
}  // namespace run
int main() { return run::main(); }

```