#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const ll MOD = 998244353; // bruh

ll pow(ll x, ll y){
    ll res = 1;
    x = x%MOD;

    while(y > 0){
        if(y&1) res = (res*x)%MOD;
        y = y >> 1;
        x = (x*x)%MOD;
    }
    return res;
}

ll inv(ll x){
    return pow(x, MOD-2);
}

vector<ll> fact{1}, invfact{1};

ll C(ll n, ll r){
    if(n < r) return 0;
    return ((fact[n]*invfact[r])%MOD*invfact[n-r])%MOD;
}

void precompute(ll n){
    for(ll i = fact.size(); i<=n; i++){
        fact.push_back((fact.back()*i)%MOD);
        invfact.push_back(inv(fact.back()));
    }
}

int main(){
    ll n, k;
    cin >> n >> k;
    precompute(n);

    vector<ll>dp(n+1,0), sum(n+1,0);
    dp[0] = sum[0] = 1;

    for(ll i=1; i<=n; i++){
        dp[i] = sum[i-1];
        if(i>k)dp[i] = (dp[i]-sum[i-k-1]+MOD)%MOD;
        dp[i] *= fact[i-1];
        dp[i] %= MOD;
        sum[i] = (sum[i-1]+(dp[i]*invfact[i])%MOD)%MOD;
    }

    cout << dp[n];
}

/*

let dp[i] = valid permutation with len i 
we can only place 1 on [1,k], assume we place on pos j
[1,j-1] can place any element 
permutation = (i-1)C(j-1) * (j-1)! 

for remain [j,n] part, it is actually a sub-problem
permutation = dp[i-j]

for(int j=0; j<=min(i,k); j++){
    dp[i] = dp[i-j]*(i-1)C(j-1)*(j-1)! 
}

dp[i-j]*(i-1)C(j-1)*(j-1)! 
= dp[i-j]*(i-1)!/(i-j)!
= (i-1)! *dp[i-j]/(i-j)!
prefix sum store


*/