Problem:
Score : 500 points

Problem Statement

There are N blocks arranged in a row. Let us paint these blocks.

We will consider two ways to paint the blocks different if and only if there is a block painted in different colors in those two ways.

Find the number of ways to paint the blocks under the following conditions:

For each block, use one of the M colors, Color 1 through Color M, to paint it. It is not mandatory to use all the colors.

There may be at most K pairs of adjacent blocks that are painted in the same color.

Since the count may be enormous, print it modulo 998244353.

Constraints

All values in input are integers.

1 \leq N, M \leq 2 \times 10^5

0 \leq K \leq N - 1

Input

Input is given from Standard Input in the following format:

N M K

Output

Print the answer.

Sample Input 1

3 2 1

Sample Output 1

6

The following ways to paint the blocks satisfy the conditions: 112, 121, 122, 211, 212, and 221. Here, digits represent the colors of the blocks.

Sample Input 2

100 100 0

Sample Output 2

73074801

Sample Input 3

60522 114575 7559

Sample Output 3

479519525

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
using namespace __gnu_pbds;
using namespace std;

typedef long long int lli;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<lli> vli;
typedef vector<pii> vii;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> OrderedSet;

const int MOD	= 998244353;
const double PI = acos(-1.0);

#define fastio			ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define sz(a) int((a).size())
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define eb emplace_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(auto &i : c)
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
#define rep(i, n)    for(int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(int i = a; i >= (n); --i)
#define fill(a)  memset(a, 0, sizeof (a))
#define error(x) cerr << #x << " = " << (x) << endl
#define errorp(p) cerr << #p << " = " << (p.first) << ", " << (p.second) << endl

template<typename T> T gcd(T a, T b){return(b?__gcd(a,b):a);}
template <typename T> T lcm(T a, T b){return (a*b)/gcd(a,b); }
int mul(int a, int b, int c){lli res=(lli)a*b;return (int)(res>=c?res%c:res);}
template<typename T>T power(T e, T n, T m){T x=1,p=e;while(n){if(n&1)x=mul(x,p,m);p=mul(p,p,m);n>>=1;}return x;}
int mod_neg(int a, int b, int c){int res;if(abs(a-b)<c)res=a-b;else res=(a-b)%c;return(res<0?res+c:res);}
template<typename T>T extended_euclid(T a, T b, T &x, T &y){T xx=0,yy=1;y=0;x=1;while(b){T q=a/b,t=b;b=a%b;a=t;\
t=xx;xx=x-q*xx;x=t;t=yy;yy=y-q*yy;y=t;}return a;}
template<typename T>T mod_inverse(T a, T n){T x,y,z=0;T d=extended_euclid(a,n,x,y);return(d>1?-1:mod_neg(x,z,n));}
template <class T>  inline void smax(T &x,T y){ x = max((x), (y));}
template <class T>  inline void smin(T &x,T y){ x = min((x), (y));}

template<long long mod = 1000000007>
struct modint{
	long long a;
 
	modint() : a(0){}
	modint(long long t){
		a = t % mod;
		if(a < 0) a += mod;
	}
 
	operator long long() const{ return a; }
 
	bool operator==(const modint &x) const{ return a == x.a; }
	bool operator!=(const modint &x) const{ return a != x.a; }
 
	modint operator-() const{ return modint(a ? (mod - a) : 0); }
	modint operator~() const{ return pow(mod - 2); }
 
	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; }
	modint operator/(const modint &x) const{ return modint(*this) /= x; }
 
	modint &operator+=(const modint &x){
		a += x.a;
		if(a >= mod) a -= mod;
		return *this;
	}
	modint &operator-=(const modint &x){
		a -= x.a;
		if(a < 0) a += mod;
		return *this;
	}
	modint &operator*=(const modint &x){
		a = a * x.a % mod;
		return *this;
	}
	modint &operator/=(const modint &x){
		a = a * (~x).a % mod; 
		return *this;
	}
 
	friend ostream &operator<<(ostream &os,const modint &x){
		return os << x.a;
	}
	friend istream &operator>>(istream &is,modint &x){
		long long t;
		is >> t;
		x = modint(t);
		return is;
	}
 
	modint pow(long long x) const{
		modint ret = 1,tmp = a;
		for(;x;tmp *= tmp,x >>= 1){
			if(x & 1) ret *= tmp;
		}
		return ret;
	}
};
 
template<class T>
struct Combination{
	vector<T> fact,factinv;
	Combination(int n) : fact(n + 1),factinv(n + 1){
		fact[0] = 1;
		for(int i = 1;i <= n;i++) fact[i] = fact[i - 1] * T(i);
		factinv[n] = ~fact[n];
		for(int i = n;i >= 1;i--) factinv[i - 1] = factinv[i] * T(i);
	}
	T nCr(int n,int r){
		if(r < 0 || n < r) return 0;
		return fact[n] * factinv[r] * factinv[n - r];
	}
};

int main(){
    #ifdef DEBUG
		freopen("test", "r", stdin);
	#endif
    fastio;
    
    int N, M, K;
    cin >> N >> M >> K;
    
    modint<MOD> ans = 0;
    Combination<modint<MOD>> comb(N + 5);
    rep(i, K + 1){
        ans += comb.nCr(N - 1, K) * modint<MOD>(M) * modint<MOD>(M - 1).pow(N - K - 1);
    }

    cout << ans << endl;

    return 0;
}