#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
inline int addmod(int x) { return x >= mod ? x - mod : x; }
inline int submod(int x) { return x < 0 ? x + mod : x; }
int fpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    y /= 2;
  }
  return ans;
}
struct pt {
  vector<int> v;
  int len;
  pt() { len = 0; }
  void setlen(int x) {
    len = x;
    v.resize(len);
  }
};
pt operator+(const pt &x, const pt &y) {
  pt z;
  z.setlen(x.len);
  for (int i = 0; i < z.len; i++) z.v[i] = addmod(x.v[i] + y.v[i]);
  return z;
}
pt operator*(const pt &x, const pt &y) {
  pt z;
  z.setlen(x.len);
  for (int i = 0; i < z.len; i++) z.v[i] = 1ll * x.v[i] * y.v[i] % mod;
  return z;
}
pt operator*(const pt &x, const int &y) {
  pt z;
  z.setlen(x.len);
  for (int i = 0; i < z.len; i++) z.v[i] = 1ll * x.v[i] * y % mod;
  return z;
}
pt operator*(const int &x, const pt &y) { return y * x; }
int inv2;
pt fwt(pt a) {
  int lim = a.len;
  for (int k = 1; k < lim; k <<= 1)
    for (int i = 0, r = k * 2; i < lim; i += r)
      for (int j = 0; j < k; j++) {
        int x = a.v[i + j], y = a.v[i + j + k];
        a.v[i + j] = addmod(x + y);
        a.v[i + j + k] = submod(x - y);
      }
  return a;
}
pt ifwt(pt a) {
  int lim = a.len;
  for (int k = 1; k < lim; k <<= 1)
    for (int i = 0, r = k * 2; i < lim; i += r)
      for (int j = 0; j < k; j++) {
        int x = a.v[i + j], y = a.v[i + j + k];
        a.v[i + j] = 1ll * inv2 * addmod(x + y) % mod;
        a.v[i + j + k] = 1ll * inv2 * submod(x - y) % mod;
      }
  return a;
}
pt modify(pt x, int fl) {
  pt y;
  y.setlen(x.len * 2);
  for (int i = 0; i < x.len; i++) y.v[i] = x.v[i];
  if (fl == 0) {
    for (int i = 0; i < x.len; i++) y.v[i + x.len] = x.v[i];
  } else {
    for (int i = 0; i < x.len; i++) y.v[i + x.len] = addmod(mod - x.v[i]);
  }
  return y;
}
int n, k, m, d[150005], sz[150005], a[150005], fr[100005], infr[100005],
    pn = 1e5;
int tsz[150005];
pt dp[150005][17], tmp2[17], tmp3[17];
int C(int x, int y) { return 1ll * fr[x] * infr[y] % mod * infr[x - y] % mod; }
void ins(int x) {
  int u = 1;
  sz[u]++;
  for (int i = m - 1; i >= 0; i--) {
    int q = (x >> i) & 1;
    u = u * 2 + q;
    sz[u]++;
  }
}
pt getpt(int u, int x) {
  pt tmp;
  tmp.setlen(1 << (m - d[u]));
  tmp.v[0] = 1;
  for (int i = 0; i < m - d[u]; i++) {
    int fl = (x >> i) & 1, lim = (1 << i);
    if (fl == 0) {
      for (int i = lim; i < 2 * lim; i++) tmp.v[i] = tmp.v[i - lim];
    } else {
      for (int i = lim; i < 2 * lim; i++)
        tmp.v[i] = addmod(mod - tmp.v[i - lim]);
    }
  }
  return tmp;
}
void add(int u, int x) {
  for (int i = 0; i <= k; i++) tmp2[i] = infr[i] * getpt(u, x - i);
  for (int i = 0; i <= k; i++) {
    tmp3[i] = dp[u][i];
    dp[u][i] = pt();
    dp[u][i].setlen(1 << (m - d[u]));
  }
  for (int i = 0; i <= k; i++)
    for (int j = 0; i + j <= k; j++) {
      int *p = &tmp2[i].v[0], *q = &tmp3[j].v[0], *ans = &dp[u][i + j].v[0];
      for (int ll = 0; ll < tmp2[i].len; ll++)
        ans[ll] = addmod(ans[ll] + 1ll * p[ll] * q[ll] % mod);
    }
}
int main() {
  inv2 = fpow(2, mod - 2);
  fr[0] = infr[0] = 1;
  for (int i = 1; i <= pn; i++) {
    fr[i] = 1ll * fr[i - 1] * i % mod;
    infr[i] = fpow(fr[i], mod - 2);
  }
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ins(a[i]);
  }
  sort(a + 1, a + n + 1);
  d[1] = 0;
  for (int i = 2; i < (1 << (m + 1)); i++) d[i] = d[i >> 1] + 1;
  for (int i = (1 << (m + 1)) - 1; i >= 1; i--) {
    if (!sz[i] || (1 << (m - d[i])) <= k) continue;
    int l, r;
    l = ((i - (1 << d[i])) << (m - d[i])), r = l + (1 << (m - d[i]));
    l += k;
    l = lower_bound(a + 1, a + n + 1, l) - a;
    r = lower_bound(a + 1, a + n + 1, r) - a;
    tsz[i] = r - l;
    if (l == r) continue;
    if ((1 << (m - d[i])) <= 2 * k) {
      l = ((i - (1 << d[i])) << (m - d[i])), r = l + (1 << (m - d[i]));
      l += k;
      for (int j = 0; j <= k; j++) {
        dp[i][j].setlen((1 << (m - d[i])));
        if (j == 0) {
          for (int ll = 0; ll < dp[i][j].len; ll++) dp[i][j].v[ll] = 1;
        }
      }
    } else {
      for (int j = 0; j <= k; j++) dp[i][j].setlen(1 << (m - d[i]));
      for (int j = 0; j <= k; j++) {
        dp[i * 2][j] = modify(dp[i * 2][j], 0);
        dp[i * 2 + 1][j] = modify(dp[i * 2 + 1][j], tsz[i * 2 + 1] % 2);
      }
      if (dp[i * 2][0].len == 0) {
        int ls = i * 2;
        for (int j = 0; j <= k; j++) {
          dp[ls][j].setlen((1 << (m - d[i])));
          if (j == 0) {
            for (int ll = 0; ll < dp[ls][j].len; ll++) dp[ls][j].v[ll] = 1;
          }
        }
      }
      if (dp[i * 2 + 1][0].len == 0) {
        int rs = i * 2 + 1;
        for (int j = 0; j <= k; j++) {
          dp[rs][j].setlen((1 << (m - d[i])));
          if (j == 0) {
            for (int ll = 0; ll < dp[rs][j].len; ll++) dp[rs][j].v[ll] = 1;
          }
        }
      }
      for (int j = 0; j <= k; j++)
        for (int ll = 0; ll + j <= k; ll++) {
          dp[i][ll + j] = dp[i][ll + j] + dp[i * 2][j] * dp[i * 2 + 1][ll];
        }
      l = (i * 2 + 1 - (1 << d[i * 2 + 1])) << (m - d[i * 2 + 1]);
      r = l + k;
    }
    l = lower_bound(a + 1, a + n + 1, l) - a;
    r = lower_bound(a + 1, a + n + 1, r) - a;
    for (int j = l; j < r; j++) add(i, a[j]);
  }
  pt p = dp[1][k];
  p = ifwt(p);
  int mul = fpow(n, 1ll * k * (mod - 2) % (mod - 1));
  for (int i = 0; i < (1 << m); i++) {
    p.v[i] = 1ll * p.v[i] * mul % mod * fr[k] % mod;
    printf("%d ", p.v[i]);
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to 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;
const int mod = 998244353;
inline int addmod(int x) { return x >= mod ? x - mod : x; }
inline int submod(int x) { return x < 0 ? x + mod : x; }
int fpow(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = 1ll * ans * x % mod;
    x = 1ll * x * x % mod;
    y /= 2;
  }
  return ans;
}
struct pt {
  vector<int> v;
  int len;
  pt() { len = 0; }
  void setlen(int x) {
    len = x;
    v.resize(len);
  }
};
pt operator+(const pt &x, const pt &y) {
  pt z;
  z.setlen(x.len);
  for (int i = 0; i < z.len; i++) z.v[i] = addmod(x.v[i] + y.v[i]);
  return z;
}
pt operator*(const pt &x, const pt &y) {
  pt z;
  z.setlen(x.len);
  for (int i = 0; i < z.len; i++) z.v[i] = 1ll * x.v[i] * y.v[i] % mod;
  return z;
}
pt operator*(const pt &x, const int &y) {
  pt z;
  z.setlen(x.len);
  for (int i = 0; i < z.len; i++) z.v[i] = 1ll * x.v[i] * y % mod;
  return z;
}
pt operator*(const int &x, const pt &y) { return y * x; }
int inv2;
pt fwt(pt a) {
  int lim = a.len;
  for (int k = 1; k < lim; k <<= 1)
    for (int i = 0, r = k * 2; i < lim; i += r)
      for (int j = 0; j < k; j++) {
        int x = a.v[i + j], y = a.v[i + j + k];
        a.v[i + j] = addmod(x + y);
        a.v[i + j + k] = submod(x - y);
      }
  return a;
}
pt ifwt(pt a) {
  int lim = a.len;
  for (int k = 1; k < lim; k <<= 1)
    for (int i = 0, r = k * 2; i < lim; i += r)
      for (int j = 0; j < k; j++) {
        int x = a.v[i + j], y = a.v[i + j + k];
        a.v[i + j] = 1ll * inv2 * addmod(x + y) % mod;
        a.v[i + j + k] = 1ll * inv2 * submod(x - y) % mod;
      }
  return a;
}
pt modify(pt x, int fl) {
  pt y;
  y.setlen(x.len * 2);
  for (int i = 0; i < x.len; i++) y.v[i] = x.v[i];
  if (fl == 0) {
    for (int i = 0; i < x.len; i++) y.v[i + x.len] = x.v[i];
  } else {
    for (int i = 0; i < x.len; i++) y.v[i + x.len] = addmod(mod - x.v[i]);
  }
  return y;
}
int n, k, m, d[150005], sz[150005], a[150005], fr[100005], infr[100005],
    pn = 1e5;
int tsz[150005];
pt dp[150005][17], tmp2[17], tmp3[17];
int C(int x, int y) { return 1ll * fr[x] * infr[y] % mod * infr[x - y] % mod; }
void ins(int x) {
  int u = 1;
  sz[u]++;
  for (int i = m - 1; i >= 0; i--) {
    int q = (x >> i) & 1;
    u = u * 2 + q;
    sz[u]++;
  }
}
pt getpt(int u, int x) {
  pt tmp;
  tmp.setlen(1 << (m - d[u]));
  tmp.v[0] = 1;
  for (int i = 0; i < m - d[u]; i++) {
    int fl = (x >> i) & 1, lim = (1 << i);
    if (fl == 0) {
      for (int i = lim; i < 2 * lim; i++) tmp.v[i] = tmp.v[i - lim];
    } else {
      for (int i = lim; i < 2 * lim; i++)
        tmp.v[i] = addmod(mod - tmp.v[i - lim]);
    }
  }
  return tmp;
}
void add(int u, int x) {
  for (int i = 0; i <= k; i++) tmp2[i] = infr[i] * getpt(u, x - i);
  for (int i = 0; i <= k; i++) {
    tmp3[i] = dp[u][i];
    dp[u][i] = pt();
    dp[u][i].setlen(1 << (m - d[u]));
  }
  for (int i = 0; i <= k; i++)
    for (int j = 0; i + j <= k; j++) {
      int *p = &tmp2[i].v[0], *q = &tmp3[j].v[0], *ans = &dp[u][i + j].v[0];
      for (int ll = 0; ll < tmp2[i].len; ll++)
        ans[ll] = addmod(ans[ll] + 1ll * p[ll] * q[ll] % mod);
    }
}
int main() {
  inv2 = fpow(2, mod - 2);
  fr[0] = infr[0] = 1;
  for (int i = 1; i <= pn; i++) {
    fr[i] = 1ll * fr[i - 1] * i % mod;
    infr[i] = fpow(fr[i], mod - 2);
  }
  scanf("%d%d%d", &n, &k, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ins(a[i]);
  }
  sort(a + 1, a + n + 1);
  d[1] = 0;
  for (int i = 2; i < (1 << (m + 1)); i++) d[i] = d[i >> 1] + 1;
  for (int i = (1 << (m + 1)) - 1; i >= 1; i--) {
    if (!sz[i] || (1 << (m - d[i])) <= k) continue;
    int l, r;
    l = ((i - (1 << d[i])) << (m - d[i])), r = l + (1 << (m - d[i]));
    l += k;
    l = lower_bound(a + 1, a + n + 1, l) - a;
    r = lower_bound(a + 1, a + n + 1, r) - a;
    tsz[i] = r - l;
    if (l == r) continue;
    if ((1 << (m - d[i])) <= 2 * k) {
      l = ((i - (1 << d[i])) << (m - d[i])), r = l + (1 << (m - d[i]));
      l += k;
      for (int j = 0; j <= k; j++) {
        dp[i][j].setlen((1 << (m - d[i])));
        if (j == 0) {
          for (int ll = 0; ll < dp[i][j].len; ll++) dp[i][j].v[ll] = 1;
        }
      }
    } else {
      for (int j = 0; j <= k; j++) dp[i][j].setlen(1 << (m - d[i]));
      for (int j = 0; j <= k; j++) {
        dp[i * 2][j] = modify(dp[i * 2][j], 0);
        dp[i * 2 + 1][j] = modify(dp[i * 2 + 1][j], tsz[i * 2 + 1] % 2);
      }
      if (dp[i * 2][0].len == 0) {
        int ls = i * 2;
        for (int j = 0; j <= k; j++) {
          dp[ls][j].setlen((1 << (m - d[i])));
          if (j == 0) {
            for (int ll = 0; ll < dp[ls][j].len; ll++) dp[ls][j].v[ll] = 1;
          }
        }
      }
      if (dp[i * 2 + 1][0].len == 0) {
        int rs = i * 2 + 1;
        for (int j = 0; j <= k; j++) {
          dp[rs][j].setlen((1 << (m - d[i])));
          if (j == 0) {
            for (int ll = 0; ll < dp[rs][j].len; ll++) dp[rs][j].v[ll] = 1;
          }
        }
      }
      for (int j = 0; j <= k; j++)
        for (int ll = 0; ll + j <= k; ll++) {
          dp[i][ll + j] = dp[i][ll + j] + dp[i * 2][j] * dp[i * 2 + 1][ll];
        }
      l = (i * 2 + 1 - (1 << d[i * 2 + 1])) << (m - d[i * 2 + 1]);
      r = l + k;
    }
    l = lower_bound(a + 1, a + n + 1, l) - a;
    r = lower_bound(a + 1, a + n + 1, r) - a;
    for (int j = l; j < r; j++) add(i, a[j]);
  }
  pt p = dp[1][k];
  p = ifwt(p);
  int mul = fpow(n, 1ll * k * (mod - 2) % (mod - 1));
  for (int i = 0; i < (1 << m); i++) {
    p.v[i] = 1ll * p.v[i] * mul % mod * fr[k] % mod;
    printf("%d ", p.v[i]);
  }
  return 0;
}

```