Problem:
Score : 700 points

Problem Statement

Takahashi throws N dice, each having K sides with all integers from 1 to K. The dice are NOT pairwise distinguishable.
For each i=2,3,...,2K, find the following value modulo 998244353:

The number of combinations of N sides shown by the dice such that the sum of no two different sides is i.

Note that the dice are NOT distinguishable, that is, two combinations are considered different when there exists an integer k such that the number of dice showing k is different in those two.

Constraints

1 \leq K \leq 2000

2 \leq N \leq 2000

K and N are integers.

Input

Input is given from Standard Input in the following format:

K N

Output

Print 2K-1 integers. The t-th of them (1\leq t\leq 2K-1) should be the answer for i=t+1.

Sample Input 1

3 3

Sample Output 1

7
7
4
7
7

For i=2, the combinations (1,2,2),(1,2,3),(1,3,3),(2,2,2),(2,2,3),(2,3,3),(3,3,3) satisfy the condition, so the answer is 7.

For i=3, the combinations (1,1,1),(1,1,3),(1,3,3),(2,2,2),(2,2,3),(2,3,3),(3,3,3) satisfy the condition, so the answer is 7.

For i=4, the combinations (1,1,1),(1,1,2),(2,3,3),(3,3,3) satisfy the condition, so the answer is 4.

Sample Input 2

4 5

Sample Output 2

36
36
20
20
20
36
36

Sample Input 3

6 1000

Sample Output 3

149393349
149393349
668669001
668669001
4000002
4000002
4000002
668669001
668669001
149393349
149393349

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#define _USE_MATH_DEFINES
#include<iostream>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<stack>
using namespace std;
#define MOD 1000000007
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define ll long long
#define ull unsigned long long
#define all(hoge) (hoge).begin(),(hoge).end()
typedef pair<ll, ll> P;
const long long INF = 1LL << 60;
typedef vector<ll> Array;
typedef vector<Array> Matrix;


//priority_queue<ll> max;//大きい順
//priority_queue<ll, Array, greater<ll>> min;//小さい順

/*firstについては昇順 secondについては降順
sort(all(wh), [&](P x, P y) {
	if (x.first == y.first)return x.second > y.second;
	return x.first < y.first;
});
*/

template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}


//substr　文字列取り出し
//upper_bound ある値より大きい一番左のイテレータを返す、lowerは以上(setに対して使うとO(N)なので、setのメンバ関数を使う
//stoi


struct Edge {//グラフ
	ll to, cap, rev;
	Edge(ll _to, ll _cap, ll _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;
void add_edge(Graph& G, ll from, ll to, ll cap,bool revFlag,ll revCap) {//最大フロー求める Ford-fulkerson
	G[from].push_back(Edge(to, cap, (ll)G[to].size()));
	if(revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));//最小カットの場合逆辺は0にする
}
ll max_flow_dfs(Graph & G, ll v, ll t, ll f, vector<bool> & used)
{
	if (v == t)
		return f;
	used[v] = true;
	for (int i = 0; i < G[v].size(); ++i) {
		Edge& e = G[v][i];
		if (!used[e.to] && e.cap > 0) {
			ll d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
			if (d > 0) {
				e.cap -= d;
				G[e.to][e.rev].cap += d;
				return d;
			}
		}
	}
	return 0;
}
//二分グラフの最大マッチングを求めたりも出来る　また二部グラフの最大独立集合は頂点数-最大マッチングのサイズ
ll max_flow(Graph & G, ll s, ll t)
{
	ll flow = 0;
	for (;;) {
		vector<bool> used(G.size());
		REP(i, used.size())used[i] = false;
		ll f = max_flow_dfs(G, s, t, INF, used);
		if (f == 0) {
			return flow;
		}
		flow += f;
	}
}
void BellmanFord(Graph& G, ll s, Array& d,Array &negative) {//O(|E||V|)
	d.resize(G.size());
	negative.resize(G.size());
	REP(i, d.size())d[i] = INF;
	REP(i, d.size())negative[i] = false;
	d[s] = 0;
	REP(k, G.size()-2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
				}						   
			}
		}
	}
	REP(k, G.size() - 2) {
		REP(i, G.size()) {
			REP(j, G[i].size()) {
				if (d[G[i][j].to] > d[i] + G[i][j].cap) {
					d[G[i][j].to] = d[i] + G[i][j].cap;
					negative[G[i][j].to] = true;
				}
				if(negative[i]==true)negative[G[i][j].to] = true;
			}
		}
	}
}
void Dijkstra(Graph& G, ll s, Array& d) {//O(|E|log|V|)
	d.resize(G.size());
	REP(i, d.size())d[i] = INF;
	d[s] = 0;
	priority_queue<P, vector<P>, greater<P>> q;
	q.push(make_pair(0, s));
	while (!q.empty()) {
		P a = q.top();
		q.pop();
		if (d[a.second] < a.first)continue;
		REP(i, G[a.second].size()) {
			Edge e = G[a.second][i];
			if (d[e.to] > d[a.second] + e.cap) {
				d[e.to] = d[a.second] + e.cap;
				q.push(make_pair(d[e.to], e.to));
			}
		}
	}
}
void WarshallFloyd(Graph& G, Matrix& d) {//O(V^3)
	d.resize(G.size());
	REP(i, d.size())d[i].resize(G.size());
	REP(i, d.size()) {
		REP(j, d[i].size()) {
			d[i][j] = INF;
		}
	}
	REP(i, G.size()) {
		REP(j, G[i].size()) {
			d[i][G[i][j].to] = G[i][j].cap;
		}
	}
	REP(i, G.size()) {
		REP(j, G.size()) {
			REP(k, G.size()) {
				chmin(d[j][k], d[j][i] + d[i][k]);
			}
		}
	}
}


class UnionFind {
	vector<int> data;
	ll num;
public:
	UnionFind(int size) : data(size, -1), num(size) { }
	bool unionSet(int x, int y) {//xとyの集合を統合する
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		num -= (x != y);
		return x != y;
	}
	bool findSet(int x, int y) {//xとyが同じ集合か返す
		return root(x) == root(y);
	}
	int root(int x) {//xのルートを返す
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(int x) {//xの集合のサイズを返す
		return -data[root(x)];
	}
	int numSet() {//集合の数を返す
		return num;
	}
};
class SumSegTree {
private:
	
	int _sum(int a, int b, int k, int l, int r) {
		if (r <= a || b <= l)return 0;    // 交差しない
		if (a <= l && r <= b)return dat[k];   // a,l,r,bの順で完全に含まれる
		else {
			int s1 = _sum(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
			int s2 = _sum(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
			return s1 + s2;
		}
	}
public:
	int n, height;
	vector<int> dat;

	// 初期化（_nは最大要素数）
	SumSegTree(int _n) {
		n = 1;
		height = 1;
		while (n < _n) {
			n *= 2;
			height++;
		}
		dat = vector<int>(2 * n - 1);
	}

	// 場所i(0-indexed)にxを足す
	void add(int i, int x) {
		i += n - 1; // i番目の葉ノードへ
		dat[i] += x;
		while (i > 0) { // 下から上がっていく
			i = (i - 1) / 2;
			dat[i] += x;
		}
	}

	// 区間[a,b)の総和。ノードk=[l,r)に着目している。
	int sum(int a, int b) {
		return _sum(a, b, 0, 0, n);
	}
};

//約数求める //約数
void divisor(ll n, vector<ll>& ret) {
	for (ll i = 1; i * i <= n; i++) {
		if (n % i == 0) {
			ret.push_back(i);
			if (i * i != n) ret.push_back(n / i);
		}
	}
	sort(ret.begin(), ret.end());

}

vector<ll> lis_fast(const vector<ll>& a) {//最長部分増加列
	const ll n = a.size();
	vector<ll> A(n, INT_MAX);
	vector<ll> id(n);
	for (int i = 0; i < n; ++i) {
		id[i] = distance(A.begin(), lower_bound(A.begin(), A.end(), a[i]));
		A[id[i]] = a[i];
	}
	ll m = *max_element(id.begin(), id.end());
	vector<ll> b(m + 1);
	for (int i = n - 1; i >= 0; --i)
		if (id[i] == m) b[m--] = a[i];
	return b;
}


ll ModPow(ll x, ll n) {
	ll res = 1LL;
	while (n > 0) {
		if (n & 1) res = res * x % MOD;
		x = x * x % MOD;
		n >>= 1;
	}
	return res;
}
//nCrとか
class Combination {
public:
	Array fact;
	Array inv;
	ll mod;
	ll mod_inv(ll x) {
		ll n = mod - 2LL;
		ll res = 1LL;
		while (n > 0) {
			if (n & 1) res = res * x % mod;
			x = x * x % mod;
			n >>= 1;
		}
		return res;
	}
	ll nCr(ll n, ll r) {
		return ((fact[n] * inv[r] % mod) * inv[n - r]) % mod;
	}
	ll nPr(ll n, ll r) {
		return (fact[n] * inv[n - r]) % mod;
	}
	ll nHr(ll n, ll r) {
		return nCr(r + n - 1, r);
	}
	Combination(ll n, ll _mod) {
		mod = _mod;
		fact.resize(n + 1);
		fact[0] = 1;
		REP(i, n) {
			fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
		}
		inv.resize(n + 1);
		REP(i, n + 1) {
			inv[i] = mod_inv(fact[i]);
		}
	}
};

bool compare_by_b(pair<ll, ll> a, pair<ll, ll> b) {//降順second
	if (a.second != b.second) {
		return a.second < b.second;
	}
	else {
		return a.first > b.first;
	}
}

bool compare_by_a(pair<ll, ll> a, pair<ll, ll> b) {//降順first
	if (a.first != b.first) {
		return a.first > b.first;
	}
	else {
		return a.second > b.second;
	}
}

ll gcd(ll m, ll n) {
	if (n == 0)return m;
	return gcd(n, m % n);
}//gcd

ll lcm(ll m, ll n) {
	return m / gcd(m, n) * n;
}


Matrix mIdentity(ll n) {
	Matrix A(n, Array(n));
	for (int i = 0; i < n; ++i) A[i][i] = 1;
	return A;
}

Matrix mMul(const Matrix& A, const Matrix& B) {
	Matrix C(A.size(), Array(B[0].size()));
	for (int i = 0; i < C.size(); ++i)
		for (int j = 0; j < C[i].size(); ++j)
			for (int k = 0; k < A[i].size(); ++k)
				(C[i][j] += (A[i][k]%MOD) * (B[k][j]%MOD))%=MOD;
	return C;
}
// O( n^3 log e )
Matrix mPow(const Matrix & A, ll e) {
	return e == 0 ? mIdentity(A.size()) : 
		e % 2 == 0 ? mPow(mMul(A, A), e / 2) : mMul(A, mPow(A, e - 1));
}


const ll mod = 998244353;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	ll k, n;
	cin >> k >> n;
	Combination comb(n + k + 1, mod);
	Array two(5000);
	two[0] = 1;
	REP(i, 4999) {
		two[i + 1] = two[i] * 2 % mod;
	}
	for (int i = 2; i <= 2 * k; i++) {
		ll ans = 0;
		if (i % 2 == 0) {
			ll num = 0;
			for (int j = 1; j <= i / 2; j++) {
				ll temp = i - j;
				if (temp <= k)num++;
			}
			num--;
			
			REP(j, num + 1) {
				if (k - 2 * num + j - 1 == 0)continue;
				(ans += ((comb.nCr(num, j) * two[j]) % mod) * comb.nHr(k - 2 * num + j - 1, n - j) % mod) %= mod;
				(ans += ((comb.nCr(num, j) * two[j]) % mod) * comb.nHr(k - 2 * num + j - 1, n - j-1) % mod) %= mod;

			}
			cout << ans << endl;
		}
		else {
			ll num = 0;
			for (int j = 1; j <= i/2;j++) {
				ll temp = i - j;
				if (temp <= k)num++;
			}
			REP(j, num+1) {
				if (k - 2 * num + j == 0)continue;
				(ans += ((comb.nCr(num, j) * two[j]) % mod) * comb.nHr(k - 2 * num + j, n - j) % mod) %= mod;
			}
			cout << ans << endl;
		}


	}

	return 0;
}