//之前的方案数定义不好用了,没法优化(反演)
//
#include<bits/stdc++.h>
using namespace std;
#define IOS {cin.tie(0);cout.tie(0);ios::sync_with_stdio(0);}
#define rep(i,j,k) for(int i=j;i<=k;++i)
#define Rep(i,j,k) for(int i=j;i<k;++i)
#define per(i,j,k) for(int i=j;i>=k;--i)
#define P pair<int,int>
#define ll long long
#define vi vector<int>
#define vl vector<ll>
const int mod=998244353,N=3e7+7;
ll fac[N],inv[N];
ll C(int i,int j){
    if(i<j)return 0;
    return fac[i]*inv[j]%mod*inv[i-j]%mod;
}
ll qp(ll x,ll y){
    ll res=1;
    while(y){
        if(y&1)res=res*x%mod;
        x=1ll*x*x%mod; y>>=1;
    }
    return res;
}
void add(ll &x,ll y){
    x+=y;if(x>=mod)x-=mod;
}
signed main(){
    IOS
    int n,d,r;cin>>n>>d>>r;
    const int M=3e7;
    fac[0]=1; rep(i,1,M)fac[i]=i*fac[i-1]%mod;
    inv[M]=qp(fac[M],mod-2); per(i,M,1)inv[i-1]=inv[i]*i%mod;
    // cout<<inv[0];return 0;
    vl h(max(n,d)+1);
    vi pri(d+1),notpri(d+1); int cnt=0;
    rep(i,2,d){
        if(!notpri[i])pri[++cnt]=i;
        for(int j=1;j<=cnt && pri[j]*i<=d;++j){
            notpri[pri[j]*i]=1;
            if(i%pri[j]==0)break;
        }
    }
    rep(i,1,d)h[i]=C(d-i+n-1,n-1);
    per(i,cnt,1){ //狄利克雷后缀和
        for(int j=d/pri[i];j;--j)add(h[j],h[j*pri[i]]); //后缀和倒着做,前缀和正着做
    }
    ll ans=0;
    rep(k,1,n){
        ll hv=h[k]*C(n,k)%mod,s;
        if(k==1)s=-1;
        else{
            s=C(k-2,r-1); if(r&1)s=-s;
        }
        if(k&1)s=-s;
        add(ans,hv*(s+mod)%mod);
    }
    cout<<(ans*fac[d]%mod*fac[n-1]%mod*qp(fac[n+d-1],mod-2)%mod+r)%mod;
}