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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define call(var) cout<<#var<<"="<<var<<endl;
#define dup(x,y) (((x)+(y)-1)/(y))
using ll=long long;
using vi = vector<int>;
using vl = vector<ll>;
using vb = vector<bool>;
using Graph = vector<vi>;
using P = pair<int,int>;
const int MOD=(int)998244353;
const int INF=(int)1e9;
const ll LINF=(ll)1e18;
template<class t,class u> inline bool chmax(t&a,u b){if(a<b){a=b;return true;}return false;}
template<class t,class u> inline bool chmin(t&a,u b){if(b<a){a=b;return true;}return false;}

struct mint {
    long long x;
    mint(long long x=0):x((x%MOD+MOD)%MOD){}
    mint operator-() const { return mint(-x);}
    mint& operator+=(const mint a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint a) {
        if ((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const mint a) {
        (x *= a.x) %= MOD;
        return *this;
    }
    mint operator+(const mint a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    // for prime MOD
    mint inv() const {
        return pow(MOD-2);
    }
    mint& operator/=(const mint a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint a) const {
        mint res(*this);
        return res/=a;
    }
    friend ostream& operator<<(ostream& os, const mint& oj);
};

ostream& operator<<(ostream& os, const mint& oj){
    os << oj.x;
    return os;
}

mint factorial(int n){
    mint res = 1;
for(int i=1;i<=n;++i){
        res *= i;
    }
    return res;
}

mint nCr(int n,int r){
    mint numerator = 1;
    if(r > n - r) r = n - r;
    for(int i=n;i>n-r;--i){
        numerator *= i;
    }
    mint denominator = 1;
    for(int i=1;i<=r;++i){
        denominator *= i;
    }
    return numerator/denominator;
}

int main(){
    int n,m,k;cin >> n >> m >> k;
    mint ans = 0;
    for(int i=0;i<=k;++i){
        mint tmp = nCr(n-1, i);
        tmp *= m;
        tmp *= mint(m-1).pow(n-1-i);
        ans += tmp;
    }
    cout << ans << "\n";
}