Problem:
Score : 200 points

Problem Statement

There are N balls placed in a row.
AtCoDeer the deer is painting each of these in one of the K colors of his paint cans.
For aesthetic reasons, any two adjacent balls must be painted in different colors.

Find the number of the possible ways to paint the balls.

Constraints

1≦N≦1000

2≦K≦1000

The correct answer is at most 2^{31}-1.

Input

The input is given from Standard Input in the following format:

N K

Output

Print the number of the possible ways to paint the balls.

Sample Input 1

2 2

Sample Output 1

2

We will denote the colors by 0 and 1. There are two possible ways: we can either paint the left ball in color 0 and the right ball in color 1, or paint the left in color 1 and the right in color 0.

Sample Input 2

1 10

Sample Output 2

10

Since there is only one ball, we can use any of the ten colors to paint it. Thus, the answer is ten.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

using namespace std;
using ll =long long;
using db = double;
using vll = vector<long long>;

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
#define che(a,string) cout<<"//"<<string <<"=="<<(a)<<"//" <<endl;
#define SORT(a) sort((a).begin(),(a).end())
#define rSORT(a) reverse((a).begin(),(a).end())
#define For(i, a, b)    for(int i = (a) ; i < (b) ; ++i)
#define rep(i, n)       For(i, 0, n)
#define leftunique(a) {SORT(a);(a).erase(unique((a).begin(),(a).end()),(a).end());}
//leftuniqueは配列から同じ要素のものを取り除く 例 4 6 2 1 3 4 1 2 なら 1 2 3 4 6 になる
#define debug(x)  cout << #x << " = " << (x) << endl;
bool IsInt(double a){int b = a/1;if(a==b){return true;}else {return false;}}
bool coY() {cout <<"Yes"<<endl;}
bool coN(){cout <<"No"<<endl;}
const int mod = 1e9 + 7;
const ll INF = 1LL << 60;
//ここから書き始める

int main()
{
	cout <<setprecision(10);
	ll n,k;
	cin >> n>>k;
	ll ans ;
	//k * k-1 * k-1 *...計n個の項がある
	ans = k * pow(k-1,n-1);
	cout << ans << endl;
}