Problem: Our child likes computer science very much, especially he likes binary trees.

Consider the sequence of n distinct positive integers: c1, c2, ..., cn. The child calls a vertex-weighted rooted binary tree good if and only if for every vertex v, the weight of v is in the set {c1, c2, ..., cn}. Also our child thinks that the weight of a vertex-weighted tree is the sum of all vertices' weights.

Given an integer m, can you for all s (1 ≤ s ≤ m) calculate the number of good vertex-weighted rooted binary trees with weight s? Please, check the samples for better understanding what trees are considered different.

We only want to know the answer modulo 998244353 (7 × 17 × 223 + 1, a prime number).

Input

The first line contains two integers n, m (1 ≤ n ≤ 105; 1 ≤ m ≤ 105). The second line contains n space-separated pairwise distinct integers c1, c2, ..., cn. (1 ≤ ci ≤ 105).

Output

Print m lines, each line containing a single integer. The i-th line must contain the number of good vertex-weighted rooted binary trees whose weight exactly equal to i. Print the answers modulo 998244353 (7 × 17 × 223 + 1, a prime number).

Examples

Input

2 3
1 2


Output

1
3
9


Input

3 10
9 4 3


Output

0
0
1
1
0
2
4
2
6
15


Input

5 10
13 10 6 4 15


Output

0
0
0
1
0
1
0
2
0
5

Note

In the first example, there are 9 good vertex-weighted rooted binary trees whose weight exactly equal to 3:

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
namespace IO {
inline char ch() {
  static char buf[1 << 21], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int in() {
  int s = 0, f = 1;
  char x;
  for (x = getchar(); x < '0' || x > '9'; x = getchar())
    if (x == '-') f = -1;
  for (; x >= '0' && x <= '9'; x = getchar())
    s = (s << 1) + (s << 3) + (x & 15);
  return f == 1 ? s : -s;
}
}  // namespace IO
using namespace IO;
const int A = 3e5 + 5;
const int logA = 18;
const int mod = 998244353;
inline int add(int a, int b) { return a + b >= mod ? a + b - mod : a + b; }
inline int dec(int a, int b) { return a - b < 0 ? a - b + mod : a - b; }
inline int mul(int a, int b) { return 1ll * a * b % mod; }
inline void Add(int &a, int b) {
  a = add(a, b);
  return;
}
inline void Dec(int &a, int b) {
  a = dec(a, b);
  return;
}
inline void Mul(int &a, int b) {
  a = mul(a, b);
  return;
}
inline int power(int s, int c) {
  int ans = 1;
  while (c) {
    if (c & 1) Mul(ans, s);
    s = mul(s, s);
    c >>= 1;
  }
  return ans;
}
inline void print(vector<int> a) {
  for (int i = 0; i < a.size(); i++) printf("%d ", a[i]);
  puts("");
  return;
}
int r[A], lim;
inline void init(int len) {
  lim = 1;
  while (lim < len) lim <<= 1;
  for (int i = 1; i < lim; i++)
    r[i] = (r[i >> 1] >> 1) | ((i & 1) ? lim >> 1 : 0);
  return;
}
inline vector<int> operator+(vector<int> a, vector<int> b) {
  if (a.size() < b.size()) a.resize(b.size());
  for (int i = 0; i < b.size(); i++) Add(a[i], b[i]);
  return a;
}
inline vector<int> operator-(vector<int> a, vector<int> b) {
  if (a.size() < b.size()) a.resize(b.size());
  for (int i = 0; i < b.size(); i++) Dec(a[i], b[i]);
  return a;
}
inline vector<int> Divx(vector<int> a) {
  for (int i = 0; i < a.size(); i++) a[i] = a[i + 1];
  a.pop_back();
  return a;
}
int *wn[logA + 1], inv[A];
inline void pre_ntt() {
  for (int i = 1; i <= logA; i++) wn[i] = new int[1 << (i - 1)];
  int k = power(3, (mod - 1) / (1 << logA));
  wn[logA][0] = 1;
  for (int i = 1; i < (1 << (logA - 1)); i++)
    wn[logA][i] = mul(wn[logA][i - 1], k);
  for (int i = logA - 1; i; i--)
    for (int j = 0; j < (1 << (i - 1)); j++) wn[i][j] = wn[i + 1][j << 1];
  inv[0] = inv[1] = 1;
  for (int i = 2; i <= (1 << logA); i++)
    inv[i] = mul(mod - mod / i, inv[mod % i]);
  return;
}
inline void ntt(vector<int> &Q, int pos) {
  for (int i = 0; i < lim; i++)
    if (i < r[i]) swap(Q[r[i]], Q[i]);
  for (int mid = 1, c = 1; mid < lim; mid <<= 1, c++)
    for (int i = 0; i < lim; i += (mid << 1))
      for (int j = 0; j < mid; j++) {
        int x = Q[i + j], y = mul(Q[i + j + mid], wn[c][j]);
        Q[i + j] = add(x, y);
        Q[i + j + mid] = dec(x, y);
      }
  if (pos) return;
  int k = power(lim, mod - 2);
  reverse(Q.begin() + 1, Q.end());
  for (int i = 0; i < lim; i++) Mul(Q[i], k);
  return;
}
inline vector<int> operator*(vector<int> a, vector<int> b) {
  int len = a.size() + b.size() - 1;
  if (a.size() < 32 || b.size() < 32) {
    vector<int> c(len, 0);
    for (int i = 0; i < a.size(); i++)
      for (int j = 0; j < b.size(); j++) Add(c[i + j], mul(a[i], b[j]));
    return c;
  }
  init(len);
  a.resize(lim), b.resize(lim);
  ntt(a, 1), ntt(b, 1);
  for (int i = 0; i < lim; i++) Mul(a[i], b[i]);
  ntt(a, 0);
  a.resize(len);
  return a;
}
inline vector<int> Inv(vector<int> a, int len) {
  vector<int> b(1, power(a[0], mod - 2)), c;
  for (int deg = 2; (deg >> 1) < len; deg <<= 1) {
    c.resize(deg);
    init(deg << 1);
    for (int i = 0; i < deg; i++) c[i] = i < a.size() ? a[i] : 0;
    c.resize(lim), b.resize(lim);
    ntt(c, 1), ntt(b, 1);
    for (int i = 0; i < lim; i++) Mul(b[i], dec(2, mul(b[i], c[i])));
    ntt(b, 0);
    b.resize(deg);
  }
  b.resize(len);
  return b;
}
inline vector<int> Der(vector<int> a) {
  for (int i = 0; i < a.size(); i++) a[i] = mul(a[i + 1], i + 1);
  a.pop_back();
  return a;
}
inline vector<int> Ing(vector<int> a) {
  a.push_back(0);
  for (int i = a.size() - 1; i; i--) a[i] = mul(inv[i], a[i - 1]);
  a[0] = 0;
  return a;
}
inline vector<int> Ln(vector<int> a, int len) {
  a = Ing(Der(a) * Inv(a, len));
  a.resize(len);
  return a;
}
inline vector<int> Exp(vector<int> a, int len) {
  vector<int> b(1, 1), c;
  for (int deg = 2; deg < (len << 1); deg <<= 1) {
    c = Ln(b, deg);
    Dec(c[0], 1);
    for (int i = 0; i < deg; i++) c[i] = dec(i < a.size() ? a[i] : 0, c[i]);
    b = b * c;
    b.resize(deg);
  }
  b.resize(len);
  return b;
}
inline vector<int> Power(vector<int> a, int c, int len) {
  a = Ln(a, len);
  for (int i = 0; i < a.size(); i++) Mul(a[i], c);
  return Exp(a, len);
}
inline vector<int> Sqrt(vector<int> a, int len) {
  vector<int> b(1, 1), c;
  for (int deg = 2; deg < (len << 1); deg <<= 1) {
    b = a * Inv(b, deg) + b;
    for (int i = 0; i < deg; i++) b[i] = mul(b[i], inv[2]);
    b.resize(deg);
  }
  b.resize(len);
  return b;
}
inline void operator-=(vector<int> &a, const vector<int> &b) {
  if (a.size() < b.size()) a.resize(b.size());
  for (int i = 0; i < b.size(); i++) Dec(a[i], b[i]);
  return;
}
inline vector<int> operator/(vector<int> a, vector<int> b) {
  int len = a.size() - b.size() + 1;
  reverse(a.begin(), a.end());
  a.resize(len);
  reverse(b.begin(), b.end());
  a = a * Inv(b, len);
  a.resize(len);
  reverse(a.begin(), a.end());
  return a;
}
inline vector<int> operator%(vector<int> a, vector<int> b) {
  if (a.size() < b.size()) return a;
  a -= b * (a / b);
  a.resize(b.size() - 1);
  return a;
}
vector<int> f[A];
inline void pre_val(int p, int l, int r, int *x) {
  if (l == r) {
    f[p].push_back(dec(0, x[l]));
    f[p].push_back(1);
    return;
  }
  int mid = (l + r) >> 1;
  pre_val(p << 1, l, mid, x), pre_val(p << 1 | 1, mid + 1, r, x);
  f[p] = f[p << 1] * f[p << 1 | 1];
  return;
}
inline void Val(int p, int l, int r, vector<int> F, int *v) {
  if (l == r) {
    v[l] = F[0];
    return;
  }
  int mid = (l + r) >> 1;
  Val(p << 1, l, mid, F % f[p << 1], v),
      Val(p << 1 | 1, mid + 1, r, F % f[p << 1 | 1], v);
  return;
}
inline vector<int> solve(vector<int> a, int len) {
  for (int i = 0; i < len; i++) a[i] = dec(0, mul(4, a[i]));
  Add(a[0], 1);
  a = Sqrt(a, len);
  Add(a[0], 1);
  a = Inv(a, len);
  for (int i = 0; i < len; i++) Mul(a[i], 2);
  return a;
}
int n, m;
int val[A];
vector<int> a;
signed main() {
  pre_ntt();
  n = in(), m = in();
  for (int i = 1; i <= n; i++) val[i] = in();
  sort(val + 1, val + 1 + n);
  int pos = 1;
  m++;
  for (int i = 0; i < m; i++) {
    if (i == val[pos]) {
      a.push_back(1);
      pos++;
    } else
      a.push_back(0);
  }
  a = solve(a, m);
  for (int i = 1; i < a.size(); i++) printf("%d\n", a[i]);
  return 0;
}