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 MAXN = 300005;
const long long MOD = 998244353;
const long long g = 3;
int n, m, N;
int rev[MAXN << 2];
long long Inv;
long long wn[2][MAXN << 2];
long long power(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) (res *= a) %= MOD;
    (a *= a) %= MOD;
    b >>= 1;
  }
  return res;
}
void INIT(int l) {
  N = 1;
  while (N < l) N <<= 1;
  int half = N >> 1;
  for (int i = 0; i < half; i++) {
    wn[0][i + half] = power(g, (MOD - 1) * i / N);
    wn[1][i + half] = power((MOD + 1) / g, (MOD - 1) * i / N);
  }
  for (int i = half - 1; i >= 0; i--) {
    wn[0][i] = wn[0][i << 1];
    wn[1][i] = wn[1][i << 1];
  }
}
void init(int l) {
  N = 1;
  while (N < l) N <<= 1;
  Inv = power(N, MOD - 2);
  int half = N >> 1;
  for (int i = 1; i < N; i++) rev[i] = (rev[i >> 1] >> 1) + (i & 1) * half;
}
void NTT(vector<long long> &a, bool inv) {
  a.resize(N);
  for (int i = 0; i < N; i++)
    if (i < rev[i]) swap(a[i], a[rev[i]]);
  for (int i = 1; i < N; i <<= 1)
    for (int j = 0; j < N; j += i << 1) {
      vector<long long>::iterator l = a.begin() + j, r = a.begin() + j + i;
      long long *w = wn[inv] + i, A;
      for (int k = 0; k < i; k++, l++, r++, w++) {
        A = *r * *w;
        *r = (*l - A) % MOD;
        *l = (*l + A) % MOD;
      }
    }
  if (inv) {
    for (int i = 0; i < N; i++) (a[i] *= Inv) %= MOD;
  }
}
vector<long long> operator+(const vector<long long> &a,
                            const vector<long long> &b) {
  int l = max(a.size(), b.size());
  vector<long long> c(l);
  for (int i = 0; i < l; i++)
    c[i] = ((i < a.size() ? a[i] : 0) + (i < b.size() ? b[i] : 0)) % MOD;
  return c;
}
vector<long long> operator-(const vector<long long> &a,
                            const vector<long long> &b) {
  int l = max(a.size(), b.size());
  vector<long long> c(l);
  for (int i = 0; i < l; i++)
    c[i] = ((i < a.size() ? a[i] : 0) - (i < b.size() ? b[i] : 0)) % MOD;
  return c;
}
vector<long long> operator*(vector<long long> a, vector<long long> b) {
  int l = a.size() + b.size() - 1;
  init(l);
  vector<long long> c(N);
  NTT(a, 0);
  NTT(b, 0);
  for (int i = 0; i < N; i++) c[i] = a[i] * b[i] % MOD;
  NTT(c, 1);
  c.resize(l);
  return c;
}
vector<long long> inv(vector<long long> A) {
  int l = A.size();
  if (l == 1) {
    A[0] = power(A[0], MOD - 2);
    return A;
  }
  vector<long long> B = A;
  B.resize((l + 1) >> 1);
  B = inv(B);
  init(l << 1);
  NTT(B, 0);
  NTT(A, 0);
  for (int i = 0; i < N; i++) (B[i] *= 2 - B[i] * A[i] % MOD) %= MOD;
  NTT(B, 1);
  B.resize(l);
  return B;
}
vector<long long> sqrt(vector<long long> A) {
  int l = A.size();
  if (l == 1) return A;
  vector<long long> B = A;
  B.resize((l + 1) >> 1);
  B = sqrt(B);
  vector<long long> C = B;
  C.resize(l);
  C = inv(C);
  init(l << 1);
  NTT(B, 0);
  NTT(A, 0);
  NTT(C, 0);
  for (int i = 0; i < N; i++)
    B[i] = (B[i] + A[i] * C[i] % MOD) * ((MOD + 1) / 2) % MOD;
  NTT(B, 1);
  B.resize(l);
  return B;
}
int main() {
  INIT(300000);
  scanf("%d%d", &n, &m);
  vector<long long> A(m + 1);
  for (int c, i = 1; i <= n; i++) {
    scanf("%d", &c);
    if (c <= m) A[c] = 2;
  }
  int cnt = 0;
  for (int i = 0;; i++)
    if (A[i]) {
      cnt = i;
      break;
    }
  A.resize(m + 2 * cnt + 1);
  vector<long long> B = A;
  for (int i = 1; i <= m; i++) B[i] *= -2;
  B[0]++;
  B = sqrt(B);
  for (int i = 0; i <= m + 2 * cnt + 1; i++) B[i] = -B[i] - A[i];
  B[0]++;
  for (int i = 0; i <= m; i++) A[i] = A[i + cnt];
  A.resize(A.size() - cnt);
  A = B * inv(A);
  for (int i = 1 + cnt; i <= m + cnt; i++) printf("%lld\n", (A[i] + MOD) % MOD);
  return 0;
}