Problem

There are $ M $ type characters. Use them to create a string of length $ N $. How many strings are used that have $ K $ or more? Find too much divided by $ 998244353 $.

Here, the difference between two strings of length $ N $ is defined as follows.

* If the two strings are $ S = S_1S_2 \ ldots S_N $, $ T = T_1T_2 \ ldots T_N $, then $ i $$ (1 \ leq i \ leq N) becomes $ S_i \ neq T_i $ ) $ Exists.

Constraints

The input satisfies the following conditions.

* $ 1 \ leq M \ leq 10 ^ {18} $
* $ 1 \ leq N \ leq 10 ^ {18} $
* $ 1 \ leq K \ leq 1000 $
* All inputs are integers

Input

The input is given in the following format.


$ M $ $ N $ $ K $


$ M, N, K $ are given on one line, separated by blanks.

Output

Output the remainder of the number of characters used in a string of length $ N $ that is greater than or equal to $ K $ divided by $ 998244353 $.

Examples

Input

2 10 1


Output

1024


Input

1 1 2


Output

0


Input

5 10 3


Output

9755400


Input

7 8 3


Output

5759460


Input

1000000000000000000 1000000000000000000 1000


Output

133611974
#include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}


template<typename T,T MOD = 1000000007>
struct Mint{
  static constexpr T mod = MOD;
  T v;
  Mint():v(0){}
  Mint(signed v):v(v){}
  Mint(long long t){v=t%MOD;if(v<0) v+=MOD;}

  Mint pow(long long k){
    Mint res(1),tmp(v);
    while(k){
      if(k&1) res*=tmp;
      tmp*=tmp;
      k>>=1;
    }
    return res;
  }

  static Mint add_identity(){return Mint(0);}
  static Mint mul_identity(){return Mint(1);}

  Mint inv(){return pow(MOD-2);}

  Mint& operator+=(Mint a){v+=a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator-=(Mint a){v+=MOD-a.v;if(v>=MOD)v-=MOD;return *this;}
  Mint& operator*=(Mint a){v=1LL*v*a.v%MOD;return *this;}
  Mint& operator/=(Mint a){return (*this)*=a.inv();}

  Mint operator+(Mint a) const{return Mint(v)+=a;};
  Mint operator-(Mint a) const{return Mint(v)-=a;};
  Mint operator*(Mint a) const{return Mint(v)*=a;};
  Mint operator/(Mint a) const{return Mint(v)/=a;};

  Mint operator-() const{return v?Mint(MOD-v):Mint(v);}

  bool operator==(const Mint a)const{return v==a.v;}
  bool operator!=(const Mint a)const{return v!=a.v;}
  bool operator <(const Mint a)const{return v <a.v;}
};
template<typename T,T MOD> constexpr T Mint<T, MOD>::mod;
template<typename T,T MOD>
ostream& operator<<(ostream &os,Mint<T, MOD> m){os<<m.v;return os;}


template<typename M>
class Enumeration{
private:
  static vector<M> fact,finv,invs;
public:
  static void init(Int n){
    n=min<decltype(M::mod)>(n,M::mod-1);

    Int m=fact.size();
    if(n<m) return;

    fact.resize(n+1,1);
    finv.resize(n+1,1);
    invs.resize(n+1,1);

    if(m==0) m=1;
    for(Int i=m;i<=n;i++) fact[i]=fact[i-1]*M(i);
    finv[n]=M(1)/fact[n];
    for(Int i=n;i>=m;i--) finv[i-1]=finv[i]*M(i);
    for(Int i=m;i<=n;i++) invs[i]=finv[i]*fact[i-1];
  }

  static M Fact(Int n){
    init(n);
    return fact[n];
  }
  static M Finv(Int n){
    init(n);
    return finv[n];
  }
  static M Invs(Int n){
    init(n);
    return invs[n];
  }

  static M C(Int n,Int k){
    if(n<k||k<0) return M(0);
    init(n);
    return fact[n]*finv[n-k]*finv[k];
  }
};
template<typename M>
vector<M> Enumeration<M>::fact = vector<M>();
template<typename M>
vector<M> Enumeration<M>::finv = vector<M>();
template<typename M>
vector<M> Enumeration<M>::invs = vector<M>();

//INSERT ABOVE HERE
signed main(){
  using M = Mint<Int, 998244353>;
  using E = Enumeration<M>;
  Int m,n,k;
  cin>>m>>n>>k;

  M ans=M(m).pow(n);
  vector<M> dp(k,0);
  M res=1;
  for(Int i=1;i<k;i++){
    dp[i]=M(i).pow(n);
    for(Int j=1;j<i;j++)
      dp[i]-=E::C(i,j)*dp[j];
    res*=M(m+1-i);
    res/=M(i);
    ans-=res*dp[i];
  }
  cout<<ans<<endl;
  return 0;
}

