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:

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Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using ll = long long;
const int M = 998244353, C = 300001;
inline ll pow(ll a, ll b, ll M) {
  if (b == 0) return 1;
  ll c = 1;
  while (b > 0) {
    if (b % 2 == 1) c = (c * a) % M;
    b /= 2;
    a = (a * a) % M;
  }
  return c;
}
void butterfly(int n, int dp[]) {
  int m = n / 2, i, curcon = 1, j;
  dp[0] = 0;
  for (j = 1; j < n;) {
    for (i = 0; i < curcon; i++, j++) dp[j] = dp[j - curcon] + m;
    curcon *= 2, m /= 2;
  }
}
ll _ntt_tmp[C];
int _ntt_dp[C];
void dft(ll poly[], int n, int ln, ll w[], ll M, ll res[]) {
  int i, d, cur, beg, tzar, md;
  ll nw;
  butterfly(n, _ntt_dp);
  for (i = 0; i < n; i++) {
    if (_ntt_dp[i] < ln)
      res[i] = poly[_ntt_dp[i]];
    else
      res[i] = 0;
  }
  for (cur = 2, d = 1; cur <= n; cur *= 2, d++) {
    for (beg = 0; beg < n; beg += cur) {
      nw = 1;
      tzar = cur / 2;
      for (i = 0; i < cur; i++) {
        md = (i >= tzar ? i - tzar : i);
        _ntt_tmp[beg + i] = (res[beg + md] + nw * res[beg + tzar + md]) % M;
        nw = (nw * w[d]) % M;
      }
    }
    for (i = 0; i < n; i++) res[i] = _ntt_tmp[i];
  }
}
const int _NTT_D = 21;
ll _ntt_res1[C], _ntt_res2[C], _ntt_full_mul[C], _ntt_w[_NTT_D][_NTT_D],
    _ntt_iw[_NTT_D][_NTT_D];
int _ntt_calc_roots[_NTT_D];
void ntt_full(ll a[], int n, ll b[], int m, ll M, ll finale[]) {
  int i = 1, mn = n + m, close = 0, p0 = 1;
  for (i = 1;; i <<= 1) {
    if ((i ^ mn) == 0) break;
    if ((i ^ mn) < mn) mn ^= i, close = 1;
    p0++;
  }
  if (close == 1) mn <<= 1, p0++;
  ll inn = pow(mn, M - 2, M);
  ll root = 3;
  if (!_ntt_calc_roots[p0]) {
    ll w0 = pow(root, 17 * 7 * (1 << (23 - p0)), M);
    _ntt_w[p0][p0] = w0;
    _ntt_iw[p0][p0] = pow(w0, M - 2, M);
    for (i = p0 - 1; i >= 0; i--)
      _ntt_w[p0][i] = (_ntt_w[p0][i + 1] * _ntt_w[p0][i + 1]) % M;
    for (i = p0 - 1; i >= 0; i--)
      _ntt_iw[p0][i] = (_ntt_iw[p0][i + 1] * _ntt_iw[p0][i + 1]) % M;
    _ntt_calc_roots[p0] = true;
  }
  dft(a, mn, n, _ntt_w[p0], M, _ntt_res1);
  dft(b, mn, m, _ntt_w[p0], M, _ntt_res2);
  for (i = 0; i < mn; i++) _ntt_full_mul[i] = (_ntt_res1[i] * _ntt_res2[i]) % M;
  dft(_ntt_full_mul, mn, mn, _ntt_iw[p0], M, finale);
  for (i = 0; i < mn; i++) finale[i] = (finale[i] * inn) % M;
}
void purge(ll t1[], ll t2[], int i, int pi, ll aim[], ll curr[], int M,
           int double_size = 1) {
  int pi2 = (1 + double_size) * pi;
  int j, ln = pi + pi2, sub_ln = 2 * pi - 1, base = pi - 1;
  ntt_full(t1, pi2, t2, pi, M, curr);
  for (j = base; j < sub_ln; j++)
    aim[i + j - base] = (aim[i + j - base] + curr[j]) % M;
  for (j = 0; j < ln; j++) curr[j] = 0;
}
ll summa[C], dp[C], c[C], post_poly[C];
int powers[C];
int main() {
  int i, j, n, m, pi, si;
  ll a;
  scanf("%d %d", &n, &m);
  for (i = 0; i < n; i++) scanf("%lld", &a), c[a] = 1;
  summa[0] = 1, dp[0] = 1;
  for (i = 1; i < C; i <<= 1) {
    for (j = i; j < C; j += i) powers[j] = i;
  }
  for (i = 1; i <= m; i++) {
    pi = powers[i];
    if (i == pi) {
      purge(c + 1, summa + i - pi, i, pi, dp, post_poly, M, 0);
    }
    if (i != powers[i]) {
      purge(summa, c + 1 + i - pi, i, pi, dp, post_poly, M);
      purge(c + 1, summa + i - pi, i, pi, dp, post_poly, M);
    }
    pi = powers[i + 1];
    if (i == pi - 1) {
      purge(dp, dp + i - pi + 1, i, pi, summa, post_poly, M, 0);
    } else {
      purge(dp, dp + i - pi + 1, i, pi, summa, post_poly, M);
      purge(dp, dp + i - pi + 1, i, pi, summa, post_poly, M);
    }
    printf("%lld\n", dp[i]);
  }
  return 0;
}