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;
const int N = 1e6 + 10;
long long qpow(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
const long long mod = 998244353;
const long long G = 3;
const long long invG = qpow(G, mod - 2, mod);
const long long inv2 = qpow(2ll, mod - 2, mod);
int tr[N];
int getLen(int n) {
  int len = 1;
  while (len < (n << 1)) len <<= 1;
  for (int i = 0; i < len; i++)
    tr[i] = (tr[i >> 1] >> 1) | (i & 1 ? len >> 1 : 0);
  return len;
}
void NTT(long long *A, int len, int type) {
  for (int i = 0; i < len; i++)
    if (i < tr[i]) swap(A[i], A[tr[i]]);
  for (int i = 2; i <= len; i <<= 1) {
    int mid = i / 2;
    long long Wn = qpow(type == 1 ? G : invG, (mod - 1) / i, mod);
    for (int k = 0; k < len; k += i) {
      long long w = 1;
      for (int l = k; l < k + mid; l++) {
        long long t = w * A[l + mid] % mod;
        A[l + mid] = (A[l] - t + mod) % mod;
        A[l] = (A[l] + t) % mod;
        w = w * Wn % mod;
      }
    }
  }
  if (type == -1) {
    long long invn = qpow(len, mod - 2, mod);
    for (int i = 0; i < len; i++) A[i] = A[i] * invn % mod;
  }
}
void getInv(long long *f, long long *g, int n) {
  if (n == 1) return (void)(g[0] = qpow(f[0], mod - 2, mod));
  getInv(f, g, (n + 1) >> 1);
  int len = getLen(n);
  static long long a[N];
  for (int i = 0; i < len; ++i) a[i] = i < n ? f[i] : 0;
  NTT(a, len, 1), NTT(g, len, 1);
  for (int i = 0; i < len; i++)
    g[i] = 1LL * (2 - 1LL * a[i] * g[i] % mod + mod) % mod * g[i] % mod;
  NTT(g, len, -1);
  for (int i = n; i < len; i++) g[i] = 0;
}
void getSqrt(long long *f, long long *g, int n) {
  if (n == 1) return (void)(g[0] = 1);
  getSqrt(f, g, (n + 1) >> 1);
  int len = getLen(n);
  static long long a[N], b[N];
  for (int i = 0; i < len; ++i) a[i] = i < n ? f[i] : 0;
  getInv(g, b, n);
  NTT(a, len, 1), NTT(b, len, 1);
  for (int i = 0; i < len; i++) a[i] = 1ll * a[i] * b[i] % mod;
  NTT(a, len, -1);
  for (int i = 0; i < n; i++) g[i] = (g[i] + a[i]) % mod * inv2 % mod;
  for (int i = n; i < len; i++) g[i] = 0;
  for (int i = 0; i < len; i++) a[i] = b[i] = 0;
}
long long f[N], g[N], a[N];
int main() {
  int m, n;
  scanf("%d%d", &m, &n);
  n++;
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    f[x] = 1;
  }
  f[0] = 1;
  for (int i = 1; i < n; i++) f[i] = (mod - 4ll * f[i]) % mod;
  getSqrt(f, g, n);
  g[0] = (g[0] + 1) % mod;
  getInv(g, a, n);
  for (int i = 1; i < n; i++) printf("%lld\n", a[i] * 2 % mod);
  return 0;
}