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 JavaScript (Node.js 12.16.1) (judge verdict: Wrong Answer)

"use strict";
var input=require("fs").readFileSync("/dev/stdin","utf8");
var cin=input.split(/ |\n/),cid=0;
function next(a){return a?cin[cid++]:+cin[cid++];}
function nexts(n,a){return a?cin.slice(cid,cid+=n):cin.slice(cid,cid+=n).map(a=>+a);}
function nextm(h,w,a){var r=[],i=0;if(a)for(;i<h;i++)r.push(cin.slice(cid,cid+=w));else for(;i<h;i++)r.push(cin.slice(cid,cid+=w).map(a=>+a));return r;}
function xArray(v){var a=arguments,l=a.length,r="Array(a["+--l+"]).fill().map(x=>{return "+v+";})";while(--l)r="Array(a["+l+"]).fill().map(x=>"+r+")";return eval(r);}
var mod = 998244353;
function mul(){for(var a=arguments,r=a[0],i=a.length;--i;)r=((r>>16)*a[i]%mod*65536+(r&65535)*a[i])%mod;return r;}
var fac=[],finv=[],invs=[],pow2=[];
function fset(n){
  fac.length=n+1,fac[0]=fac[1]=1;
  for(var i=2;i<=n;)fac[i]=mul(fac[i-1],i++);
  finv.length=n+1,finv[0]=finv[1]=1,finv[n]=inv(fac[n]);
  for(i=n;2<i;)finv[i-1]=mul(finv[i],i--);
  invs.length=n+1,invs[0]=invs[1]=1;
  for(i=2;i<=n;)invs[i]=mul(fac[i-1],finv[i++]);
  //pow2.length=n+1,pow2[0]=1;
  //for(i=1;i<=n;i++)pow2[i]=pow2[i-1]*2%mod;
}
function inv(b){for(var a=mod,u=0,v=1,t;b;v=t)t=a/b|0,a-=t*b,u-=t*v,t=a,a=b,b=t,t=u,u=v;u%=mod;return u<0?u+mod:u;}
function pow(a,n){for(var r=1;n;a=mul(a,a),n>>=1)if(n&1)r=mul(a,r);return r;}
function ncr(n,r){return mul(fac[n],finv[r],finv[n-r]);}
function npr(n,r){return mul(fac[n],finv[r]);}
function mput(n){return (n%mod+mod)%mod;}

var myOut = main();
if(myOut !== undefined)console.log(myOut);

function main(){
  var [n,m,k] = nexts(3);
  var kx = mul(m,pow(m-1,n-1));
  var ans = kx;
  fset(Math.max(n,m));
  for(var i = 1; i <= k; i++){
    kx = mul(kx,invs[m-1]);
    var ki = mul(kx,ncr(n-1,i));
    ans += ki;
  }
  return ans%mod;
}