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 namespace std;
template <typename T, typename U>
inline void smin(T &a, U b) {
  if (a > b) a = b;
}
template <typename T, typename U>
inline void smax(T &a, U b) {
  if (a < b) a = b;
}
template <class T>
inline void gn(T &first) {
  char c, sg = 0;
  while (c = getchar(), (c > '9' || c < '0') && c != '-')
    ;
  for ((c == '-' ? sg = 1, c = getchar() : 0), first = 0; c >= '0' && c <= '9';
       c = getchar())
    first = (first << 1) + (first << 3) + c - '0';
  if (sg) first = -first;
}
template <class T>
inline void print(T first) {
  if (first < 0) {
    putchar('-');
    return print(-first);
  }
  if (first < 10) {
    putchar('0' + first);
    return;
  }
  print(first / 10);
  putchar(first % 10 + '0');
}
template <class T, class T1>
inline void gn(T &first, T1 &second) {
  gn(first);
  gn(second);
}
template <class T, class T1, class T2>
inline void gn(T &first, T1 &second, T2 &z) {
  gn(first);
  gn(second);
  gn(z);
}
template <class T>
inline void println(T first) {
  print(first);
  putchar('\n');
}
template <class T>
inline void printsp(T first) {
  print(first);
  putchar(' ');
}
template <class T1, class T2>
inline void print(T1 x1, T2 x2) {
  printsp(x1), println(x2);
}
template <class T1, class T2, class T3>
inline void print(T1 x1, T2 x2, T3 x3) {
  printsp(x1), printsp(x2), println(x3);
}
template <class T1, class T2, class T3, class T4>
inline void print(T1 x1, T2 x2, T3 x3, T4 x4) {
  printsp(x1), printsp(x2), printsp(x3), println(x4);
}
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
const int mod = 998244353;
namespace fft {
const int root = 31;
const int root_1 = power(root, mod - 2, mod);
int mPow(int a, int first) { return power(a, first, mod); }
void fft(vector<int> &a, bool invert) {
  int n = a.size();
  int PW = mPow(invert ? root_1 : root, (mod - 1) / n);
  for (int m = n, h; h = m / 2, m >= 2; PW = 1LL * PW * PW % mod, m = h) {
    for (int i = 0, w = 1; i < h; ++i, w = 1LL * w * PW % mod)
      for (int j = i; j < n; j += m) {
        int k = j + h, first = (a[j] - a[k] + mod) % mod;
        a[j] += a[k];
        a[j] %= mod;
        a[k] = 1LL * w * first % mod;
      }
  }
  for (int i = 0, j = 1; j < n - 1; ++j) {
    for (int k = n / 2; k > (i ^= k); k /= 2)
      ;
    if (j < i) swap(a[i], a[j]);
  }
  if (invert) {
    int rev = mPow(n, mod - 2);
    for (int i = 0; i < n; ++i) a[i] = 1LL * a[i] * rev % mod;
  }
}
vector<int> multiply(const vector<int> &a, const vector<int> &b) {
  int n = a.size() + b.size();
  vector<int> na = a, nb = b;
  int mx = 1;
  while (mx < n) mx <<= 1;
  na.resize(mx);
  nb.resize(mx);
  fft(na, false);
  fft(nb, false);
  for (int i = 0; i < mx; ++i) na[i] = 1LL * na[i] * nb[i] % mod;
  fft(na, true);
  na.resize(n);
  return na;
}
vector<int> findInverse(const vector<int> &f) {
  int n = f.size();
  vector<int> r(1, mPow(f[0], mod - 2));
  for (int it = 2; it <= n; it <<= 1) {
    vector<int> nr =
        multiply(multiply(r, r), vector<int>(f.begin(), f.begin() + it));
    nr.resize(it);
    for (int i = 0; i < nr.size(); ++i) nr[i] = mod - nr[i];
    for (int i = 0; i < it / 2; ++i) nr[i] = (2LL * r[i] + nr[i]) % mod;
    r = nr;
  }
  return r;
}
bool checkInverse(const vector<int> &a, const vector<int> &b) {
  for (int i = 0; i < a.size(); i++) {
    int val = 0;
    for (int j = 0; j <= i; j++) {
      val = (val + 1LL * a[i - j] * b[j]) % mod;
    }
    if (val != (i == 0)) return false;
  }
  return true;
}
vector<int> squareRoot(const vector<int> &f) {
  int n = f.size();
  vector<int> r(1, 1);
  int inv2 = mPow(2, mod - 2);
  for (int it = 2; it <= n; it <<= 1) {
    r.resize(it);
    vector<int> res =
        multiply(findInverse(r), vector<int>(f.begin(), f.begin() + it));
    res.resize(it);
    for (int i = 0; i < r.size(); ++i) {
      res[i] += r[i];
      if (res[i] >= mod) res[i] -= mod;
    }
    for (int i = 0; i < res.size(); ++i) res[i] = 1LL * res[i] * inv2 % mod;
    r = res;
  }
  return r;
}
};  // namespace fft
vector<int> a;
int main() {
  int n, m;
  cin >> n >> m;
  int mx = 1;
  while (mx <= m) mx <<= 1;
  a.resize(mx);
  for (int i = 0; i < n; i++) {
    int u;
    gn(u);
    if (u <= m) a[u] = mod - 4;
  }
  a[0] = 1;
  a = fft::squareRoot(a);
  a[0]++;
  a[0] %= mod;
  a = fft::findInverse(a);
  for (int i = 1; i <= m; i++) println(a[i] * 2 % mod);
  return 0;
}