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# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCorder
{
	public class Program
	{
		static void Main(string[] args)
		{
			Scanner cin = new Scanner();
			int n = cin.Int();
			int k = cin.Int();

			long ans = k;
			for (int i = 0; i < n - 1; i++) {
				ans *= k - 1;
			}

			Console.WriteLine(ans);
		}
	}

	public class Scanner
	{
		string[] s;
		int i;

		char[] cs = new char[] { ' ' };

		public Scanner()
		{
			s = new string[0];
			i = 0;
		}

		public string Next()
		{
			if (i < s.Length) return s[i++];
			string st = Console.ReadLine();
			while (st == "") st = Console.ReadLine();
			s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
			if (s.Length == 0) return Next();
			i = 0;
			return s[i++];
		}

		public int Int()
		{
			return int.Parse(Next());
		}
		public int[] ArrayInt(int N, int add = 0)
		{
			int[] Array = new int[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Int() + add;
			}
			return Array;
		}

		public long Long()
		{
			return long.Parse(Next());
		}

		public long[] ArrayLong(int N, long add = 0)
		{
			long[] Array = new long[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Long() + add;
			}
			return Array;
		}

		public double Double()
		{
			return double.Parse(Next());
		}


		public double[] ArrayDouble(int N, double add = 0)
		{
			double[] Array = new double[N];
			for (int i = 0; i < N; i++) {
				Array[i] = Double() + add;
			}
			return Array;
		}
	}
}