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;
#define REPP(i,a,b,c) for(int i=a; i<=b; i+=c)
#define REP(i,a,b) REPP(i,a,b,1)
#define REVV(i,a,b,c) for(int i=a; i>=b; i-=c)
#define REV(i,a,b) REVV(i,a,b,1)
#define FOR(i,a) REP(i,0,(int)a-1)
#define FORD(i,a) REV(i,(int)a-1,0)
#define PB push_back
#define POB pop_back
#define MP make_pair
#define FI first
#define SE second
typedef long long ll;
typedef unsigned long long ull;
typedef pair < int , int > pii;
typedef vector < int > vi;
typedef vector < pii > vii;
typedef vector < ll > vl;

const double EPS = 1e-9;
const double PI = acos(-1);
const int INF = 1e9;
const ll MOD = 1e9 + 7;

ll addMod(ll a, ll b){
	return (a+b) % MOD;
}

ll mulMod(ll a, ll b){
	return (a*b) % MOD;
}

/////////////////////////////////////////////////////////

ll n,m;

ll solve(){
	ll ret = m;
	REP(i,2,n) ret *= (m-1);
	return ret;
}

int main(){
	ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n >> m;
	cout << solve() << endl;
	return 0;
}