/*
A: 10min
B: 20min
C: 30min
D: 40min
*/ 
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
#include <sstream>
#define pb push_back 
#define all(x) (x).begin(),(x).end()
#define mem(f, x) memset(f,x,sizeof(f)) 
#define fo(i,a,n) for(int i=(a);i<=(n);++i)
#define fo_(i,a,n) for(int i=(a);i<(n);++i)
#define debug(x) cout<<#x<<":"<<x<<endl;
#define endl '\n'
using namespace std;
//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

template<typename T>
ostream& operator<<(ostream& os,const vector<T>&v){for(int i=0,j=0;i<v.size();i++,j++)if(j>=5){j=0;puts("");}else os<<v[i]<<" ";return os;}
template<typename T>
ostream& operator<<(ostream& os,const set<T>&v){for(auto c:v)os<<c<<" ";return os;}
template<typename T1,typename T2>
ostream& operator<<(ostream& os,const map<T1,T2>&v){for(auto c:v)os<<c.first<<" "<<c.second<<endl;return os;}
template<typename T>inline void rd(T &a) {
    char c = getchar(); T x = 0, f = 1; while (!isdigit(c)) {if (c == '-')f = -1; c = getchar();}
    while (isdigit(c)) {x = (x << 1) + (x << 3) + c - '0'; c = getchar();} a = f * x;
}

typedef pair<int,int>PII;
typedef pair<long,long>PLL;

typedef long long ll;
typedef unsigned long long ull; 
const int N=2e5+10,MOD=998244353;
ll n,m,_;

void solve() {
	int n, k;
	cin >> n >> k;
	vector<int> ans(n + 1);
	int sq = 2 * sqrt(n);
	vector<vector<int>> dp(2, vector<int>(n + 1)), presum(2, vector<int>(n + 1));
	/*
	dp[i][j] 表示 i 步，走到了 j 点。
	x = i+k-1 
	dp[i][j] = dp[i-1][j-x] + dp[i-1][j-2*x] + ... +;

	presum[i][j]  = dp[i][j] + dp[i][j-x] + dp[i][j-2*x] + ... +;
	*/
	dp[0][0] = 1;
	presum[0][0] = 1;
	int cur = 1;
	for (int j = 0; j < sq; j++) {
		cur ^= 1;
		for (int i = j+k-1; i <= n; i++) {
			int x = i-(j+k-1);
			dp[cur][i] = presum[cur ^ 1][x];
			presum[cur][i] = (presum[cur][i - x] + dp[cur ^ 1][i]) % MOD;
		}
		for (int i = 0; i <= n; i++) {
      		ans[i] = (dp[cur][i] + ans[i]) % MOD;
		}
	}

	for (int i = 1;i <= n; i++) {
		cout << ans[i] << " ";
	}
	cout << "\n";

}

int main(){
    solve();
	return 0;
}