积性函数可以只求质数的函数值，
优化插值过程，维护前缀积后缀积

#include<bits/stdc++.h>
using namespace std;
#define rep( i, s, t ) for( register int i = s; i <= t; ++ i )
#define drep( i, s, t ) for( register int i = t; i >= s; -- i )
#define re register
#define int long long
#define LL long long
const int N = 1e7 + 5 ;  
const int mod = 998244353 ; 
int read() {
    char cc = getchar(); int cn = 0, flus = 1;
    while(cc < '0' || cc > '9') {  if( cc == '-' ) flus = -flus;  cc = getchar();  }
    while(cc >= '0' && cc <= '9')  cn = cn * 10 + cc - '0', cc = getchar();
    return cn * flus;
}
int pr[N], sumf[2 * N], top ;
int n, b[N], pre[N], nxt[N], inv[N], K ; 
bool isp[2 * N] ; 
LL pow2( int x, int k ) {
    LL ans = 1, base = x ; 
    while( k ) {
        if( k & 1 ) ans *= base, ans %= mod ;
        base *= base, base %= mod ; 
        k >>= 1 ; 
    }
    return ans % mod ; 
}
void init( int x ) {
    sumf[1] = 1 ;
    rep( i, 2, x ) {
        if( !isp[i] ) pr[++ top] = i, sumf[i] = pow2( i, K );
        for( re int j = 1; j <= top && i * pr[j] <= x; ++ j ) {
            isp[i * pr[j]] = 1, sumf[i * pr[j]] = ( sumf[i] * sumf[pr[j]] ) % mod ; 
            if( i % pr[j] == 0 ) break; 
        } 
    }
    rep( i, 2, x ) sumf[i] = sumf[i - 1] + sumf[i] ; 
}
void lagr( int x ) {
    b[1] = 0, nxt[0] = nxt[x + 1] = pre[x + 1] = pre[0] = inv[0] = 1; int Fac = 1 ; 
    rep( i, 2, x ) b[i] = ( b[i - 1] + sumf[2 * i - 1] - sumf[i] + mod ) % mod ; 
    rep( i, 1, x ) Fac = ( Fac * i ) % mod, pre[i] = ( n - i ) * pre[i - 1] % mod;
    inv[x] = pow2( Fac, mod - 2 ) ; 
    drep( i, 1, x ) nxt[i] = ( nxt[i + 1] * ( n - i ) ) % mod, inv[i - 1] = ( inv[i] * i ) % mod ; 

}
int get( int x ) {
    int Ans = 0, d ;
    rep( i, 1, x ) {
        d = ( b[i] * pre[i - 1] ) % mod, d = ( d * nxt[i + 1] ) % mod;
        d = ( d * ( ( inv[i - 1] * inv[x - i] ) % mod ) ) % mod ; 
        if( ( x - i ) % 2 ) d = mod - d ; 
        Ans = ( Ans + d ) % mod ;
    }
    return Ans % mod ; 
}
signed main()
{
    n = read(), K = read() ; 
    init( 2 * K + 6 ), lagr( K + 3 ) ; 
    int Kth = pow2( n, mod - 2 ) * 2 % mod ; 
    printf("%d\n", ( 1ll * get( K + 3 ) * Kth % mod ) ) ; 
    return 0;
}