C. Colorful Bricks 组合数学

C. Colorful Bricks

http://codeforces.com/problemset/problem/1081/C

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

On his free time, Chouti likes doing some housework. He has got one new task, paint some bricks in the yard.

There are nn bricks lined in a row on the ground. Chouti has got mm paint buckets of different colors at hand, so he painted each brick in one of those mm colors.

Having finished painting all bricks, Chouti was satisfied. He stood back and decided to find something fun with these bricks. After some counting, he found there are kk bricks with a color different from the color of the brick on its left (the first brick is not counted, for sure).

So as usual, he needs your help in counting how many ways could he paint the bricks. Two ways of painting bricks are different if there is at least one brick painted in different colors in these two ways. Because the answer might be quite big, you only need to output the number of ways modulo 998244353998244353.

Input

The first and only line contains three integers nn, mm and kk (1≤n,m≤2000,0≤k≤n−11≤n,m≤2000,0≤k≤n−1) — the number of bricks, the number of colors, and the number of bricks, such that its color differs from the color of brick to the left of it.

Output

Print one integer — the number of ways to color bricks modulo 998244353998244353.

Examples

input

Copy

3 3 0

output

Copy

3

input

Copy

3 2 1

output

Copy

4

Note

In the first example, since k=0k=0, the color of every brick should be the same, so there will be exactly m=3m=3 ways to color the bricks.

In the second example, suppose the two colors in the buckets are yellow and lime, the following image shows all 44 possible colorings.

题意：给你n个砖头和m中颜料和一个k，k是代表着恰好有k个砖头与他本身左边的颜色不同。

做法：

我们可以在n-1块方块选择k个方块，来涂色，满足其与左边不同，即是,

我们来解释下，先选出k个，首先第一个可以取m种颜色，按照题意 k个方块都能取m-1种颜色。

假设我们选好了第一块的颜色，若下一个不是 不属于k中，其颜色一定与他左边的那块相同，若下一个是k中的其中一个，我们要满足与他左边那块不同，只能有m-1中颜色可以选择。以此类推。
来自：https://blog.csdn.net/ljd201724114126/article/details/85055904

#include
using namespace std;
typedef long long ll;
const ll mod=998244353 ;
ll powmod(ll a,ll b)
{
	ll res=1;
	for(;b;b>>=1){if(b&1) res=res*a%mod;a=a*a%mod;}
	return res%mod;
}
ll f(ll x)
{
	ll res=1;
	for(ll i=1;i<=x;i++) res=res*i%mod;
	return res%mod;
}
ll inv(ll v)
{
	return powmod(v,mod-2);
}
ll getC(ll a,ll b)
{
	return f(a)*inv(f(a-b)*f(b)%mod)%mod;
}
int main()
{
	ll n,m,k;
	cin>>n>>m>>k;
	printf("%lld\n",((getC(n-1,k)*m%mod)*powmod(m-1,k))%mod);
}