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>
inline int mod(int x) { return x >= 998244353 ? x - 998244353 : x; }
inline int calc(int x, int y) {
  int z = 1;
  for (; y; x = (long long)x * x % 998244353, y >>= 1)
    if (y & 1) z = (long long)x * z % 998244353;
  return z;
}
char G[1100000], *H = G;
inline int read() {
  int x = 0;
  while (*H < 48 || 57 < *H) H++;
  while (47 < *H && *H < 58) x = (x << 1) + (x << 3) + (*H++ ^ 48);
  return x;
}
int S[10], T;
inline void print(int x) {
  int y;
  do y = x / 10, S[T++] = x - y * 10, x = y;
  while (x);
  while (T) *H++ = S[--T] ^ 48;
}
int I, J, f[262144], i, m, u[19], v, w[262144 | 1], x, y, z;
inline void NTT(int* a, int n, int s) {
  for (i = f[n >> 1], I = J = 0; J < n; I += i, J++)
    if (f[I] > J) z = a[f[I]], a[f[I]] = a[J], a[J] = z;
  for (i = 1, z = 0; i < n; i <<= 1, z++)
    for (I = 0, J = i, s >>= 1; J < n; I += i, J += i)
      for (v = m & s; J & i; v += s)
        x = a[I], y = (long long)a[J] * w[v] % 998244353, a[I++] = mod(x + y),
        a[J++] = mod(998244353 + x - y);
  if (s < 0)
    for (i = 0, z = u[z]; i < n; i++) a[i] = (long long)a[i] * z % 998244353;
}
int B[262144], C[262144];
inline void inv(int* A, int n) {
  for (int j = *B = 1; j < n; memset(B + j, 0, j << 2)) {
    j <<= 1, memcpy(C, A, j << 2), NTT(B, j << 1, m), NTT(C, j << 1, m);
    for (i = j << 1; i--;)
      B[i] = (2 + (long long)(998244353 - C[i]) * B[i]) % 998244353 * B[i] %
             998244353;
    NTT(B, j << 1, -m);
  }
  memset(C, 0, n << 3);
}
int D[262144];
inline void sqrt(int* A, int n) {
  for (int j = *D = 1; j < n;
       memset(B, 0, j << 3), memset(C, 0, j << 3), memset(D + j, 0, j << 2)) {
    inv(D, j <<= 1), memcpy(C, A, j << 2), NTT(B, j << 1, m), NTT(C, j << 1, m);
    for (i = j << 1; i--;)
      B[i] = 499122177ll * B[i] % 998244353 * C[i] % 998244353;
    NTT(B, j << 1, -m), memcpy(D + (j >> 1), B + (j >> 1), j << 1);
  }
}
int A[262144], n, q, t;
int main() {
  fread(G, 1, 1100000, stdin), n = read(), q = read(), m = 1, *A = *u = 1;
  while (m <= q << 1) m <<= 1, t++;
  for (i = 1; i <= t; i++) u[i] = 499122177ll * u[i - 1] % 998244353;
  w[*w = w[m] = 1] = calc(3, 998244353 >> t--), f[1] = 1 << t;
  for (i = 2; i < m; i++)
    f[i] = f[i >> 1] >> 1 | (i & 1) << t,
    w[i] = (long long)w[i - 1] * w[1] % 998244353;
  while (n--) A[read()] = 998244353 - 4;
  sqrt(A, m >> 1), ++*D;
  for (i = 0; i <= q; i++) A[i] = 499122177ll * D[i] % 998244353;
  inv(A, m >> 1), H = G;
  for (i = 1; i <= q; i++) print(B[i]), *H++ = '\n';
  return 0 * fwrite(G, 1, H - G, stdout);
}