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;
typedef pair<int,int> P;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define INF 2000000000
#define LLINF 1000000000000000ll
#define sz(x) ((int)(x).size())
#define fi first
#define sec second
#define all(x) (x).begin(),(x).end()
#define sq(x) ((x)*(x))
#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define repn(i,a,n) for(int (i)=(a);(i)<(int)(n);(i)++)
#define EQ(a,b) (abs((a)-(b))<eps)
#define dmp(x) cerr << __LINE__ << " " << #x << " " << x << endl;
template<class T> void chmin(T& a,const T& b){if(a>b)a=b;}
template<class T> void chmax(T& a,const T& b){if(a<b)a=b;}
template<class T,class U>
ostream& operator << (ostream& os,pair<T,U>& p){
	os << p.fi << ',' << p.sec; return os;
}
template<class T,class U>
istream& operator >> (istream& is,pair<T,U>& p){
	is >> p.fi >> p.sec; return is;
}
template<class T>
ostream& operator << (ostream &os,const vector<T> &vec){
	for(int i=0;i<vec.size();i++){
		os << vec[i];
		if(i+1<vec.size())os << ' ';
	}
	return os;
}
template<class T>
istream& operator >> (istream &is,vector<T>& vec){
	for(int i=0;i<vec.size();i++)is >> vec[i];
	return is;
}
void fastio(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	cout<<fixed<<setprecision(20);
}
// How to devide :
// ModInt a(6ll);
// ModInt b(2ll);
// a *= b.exp(MOD-2ll);  -> a/=b;  result: a = 3
ll MOD = 998244353ll; // if inv is needed, this shold be prime.
struct ModInt{
	ll val;
	ModInt():val(0ll){}
	ModInt(ll v):val(((v%MOD)+MOD)%MOD){}
	ModInt exp(ll y)const{
		if(!y)return ModInt(1ll);
		ModInt a = exp(y/2ll);
		a *= a;
		if(y&1)a*=(*this);
		return a;
	}
	bool operator==(const ModInt& x)const{return val==x.val;}
	inline bool operator!=(const ModInt& x)const{return !(*this==x);}
	bool operator<(const ModInt& x)const{return val<x.val;}
	bool operator>(const ModInt& x)const{return val>x.val;}
	inline bool operator>=(const ModInt& x)const{return !(*this<x);}
	inline bool operator<=(const ModInt& x)const{return !(*this>x);}
	ModInt& operator+=(const ModInt& x){if((val+=x.val)>=MOD)val-=MOD;return *this;}
	ModInt& operator-=(const ModInt& x){if((val+=MOD-x.val)>=MOD)val-=MOD;return *this;}
	ModInt& operator*=(const ModInt& x){(val*=x.val)%=MOD;return *this;}
	ModInt operator+(const ModInt& x)const{return ModInt(*this)+=x;}
	ModInt operator-(const ModInt& x)const{return ModInt(*this)-=x;}
	ModInt operator*(const ModInt& x)const{return ModInt(*this)*=x;}
};
istream& operator>>(istream&i,ModInt&x){i>>x.val;return i;}
ostream& operator<<(ostream&o,const ModInt&x){o<<x.val;return o;}
ModInt pow(ModInt a,ll x){
	ModInt res = ModInt(1ll);
	while(x){
		if(x&1ll)res *= a;
		x >>= 1;
		a = a*a;
	}
	return res;
}
const int SIZE = 100100;
ModInt inv[SIZE+10],fac[SIZE+10],facinv[SIZE+10];
// notice: 0C0 = 1 
ModInt nCr(int n,int r){
	assert(!(n<r));
	assert(!(n<0||r<0));
	return fac[n]*facinv[r]*facinv[n-r];
}
void init(){
	fac[0]=ModInt(1ll);
	for(int i=1;i<=SIZE;i++)fac[i]=fac[i-1]*ModInt(i);
	inv[1]=ModInt(1ll);
	for(int i=2;i<=SIZE;i++)inv[i]=ModInt(0ll)-ModInt(MOD/i)*inv[MOD%i];
	facinv[0]=ModInt(1ll);
	for(int i=1;i<=SIZE;i++)facinv[i]=facinv[i-1]*inv[i];
	return;
}
ModInt nCk(ll n,ll k){
  ModInt ret = ModInt(1ll);
  for(int i=0;i<k;i++){
    ret *= ModInt(n-i);
  }
  for(int i=1;i<=k;i++){
    ret *= pow(ModInt(i),MOD-2);
  }
  return ret;
}
ll M,N,K;
ModInt y[100100];
ModInt a[100100];
ModInt b[100100];
int main(){
  cin >> M >> N >> K;
  ModInt ans = pow(ModInt(M),N);
  a[0] = ModInt(M);
  for(int i=1;i<=3*K;i++){
    a[i] = a[i-1]*ModInt(M-i);
  }
  b[0] = ModInt(1ll);
  for(int i=1;i<=3*K;i++){
    b[i] = b[i-1]*ModInt(i);
  }
  for(int i=1;i<K;i++){
    if(i>M)break;
    y[i] = nCk(M,i)*pow(ModInt(i),N);
    for(int j=1;j<i;j++){
      y[i]-=y[j]*a[i-1]*pow(a[j-1],MOD-2ll)*pow(b[i-j],MOD-2ll);
    }
    // cout << y[i] << endl;
    ans -= y[i];
  }
  cout << ans << endl;
	return 0;
}

