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

long long mod = 998244353;
//入力が必ずmod未満の時に使う.
struct mint{
    long long v = 0;
    mint(){} mint(int a){v = a;} mint(long long a){v = a;} mint(unsigned long long a){v = a;}
    long long val(){return v;}
    void modu(){v %= mod;}
 
    mint repeat2mint(long long a,long long b){
        mint ret = 1,p = a;
        while(b){if(b&1) ret *= p; p *= p; b >>= 1;}
        return ret;
    };
 
    mint operator-(){return mint(0)-mint(v);}
    mint operator+(mint b){return (v+b.v)%mod;}
    mint operator-(mint b){return (v-b.v+mod)%mod;}
    mint operator*(mint b){return v*b.v%mod;}
    mint operator/(mint b){
        if(b.v == 0) assert(false);
        return v*(repeat2mint(b.v,mod-2)).v%mod;
    }
 
    void operator+=(mint b){v = (v+b.v)%mod;}
    void operator-=(mint b){v = (v-b.v+mod)%mod;}
    void operator*=(mint b){v = v*b.v%mod;}
    void operator/=(mint b){
    if(b.v == 0) assert(false);
        v = v*repeat2mint(b.v,mod-2).v%mod;
    }
 
    void operator++(int){v = (v+1)%mod; return;}
    void operator--(int){v = (v-1+mod)%mod; return;}
    bool operator==(mint b){return v == b.v;}
    bool operator!=(mint b){return v != b.v;}
    bool operator>(mint b) {return v >  b.v;}
    bool operator>=(mint b){return v >= b.v;}
    bool operator<(mint b) {return v <  b.v;}
    bool operator<=(mint b){return v <= b.v;}
 
    mint pow(long long x){return repeat2mint(v,x);}
    mint inv(){return mint(1)/v;}
};			

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int N,K; cin >> N >> K;
/*     vector<int> P(N);
    iota(P.begin(),P.end(),1);
    vector<int> diff(N);
    do{
        for(int i=0; i<N-1; i++) diff.at(abs(P.at(i+1)-P.at(i)))++;
    }while(next_permutation(P.begin(),P.end()));
    for(auto d : diff) cout << d << " " ; cout << endl; */
    N--;
    if(N >= mod){cout << 0 << endl; return 0;}

    vector<mint> ume = {1,295201906,160030060,957629942,545208507,213689172,760025067,939830261,506268060,39806322,808258749,440133909,686156489,741797144,390377694,12629586,544711799,104121967,495867250,421290700,117153405,57084755,202713771,675932866,79781699,956276337,652678397,35212756,655645460,468129309,761699708,533047427,287671032,206068022,50865043,144980423,111276893,259415897,444094191,593907889,573994984,892454686,566073550,128761001,888483202,251718753,548033568,428105027,742756734,546182474,62402409,102052166,826426395,159186619,926316039,176055335,51568171,414163604,604947226,681666415,511621808,924112080,265769800,955559118,763148293,472709375,19536133,860830935,290471030,851685235,242726978,169855231,612759169,599797734,961628039,953297493,62806842,37844313,909741023,689361523,887890124,380694152,669317759,367270918,806951470,843736533,377403437,945260111,786127243,80918046,875880304,364983542,623250998,598764068,804930040,24257676,214821357,791011898,954947696,183092975};
    mint now = ume.at(N/(int)1e7)*2;
    for(int i=N-N%(int)1e7+1; i<=N; i++) now *= i;
    N++;
    mint now2 = 0;
    for(int i=1; i<=N; i++) now2 += mint(N-i).pow(K)*i;
    cout << (now*now2).v << endl; 
}