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.Text;
using System.Linq;
using System.Collections;
using System.Threading.Tasks;
using System.Collections.Generic;
using System.Runtime.InteropServices;
using static System.Console;
using static System.Math;

namespace NotFounds
{
    public class Program
    {
        public static void Main(string[] args)
        {
            new Program().Solve();
        }

        public void Solve()
        {
            MyInputStream cin = new MyInputStream();

            var n = cin.ReadInt();
            var k = cin.ReadInt();

            WriteLine(k * Pow(k - 1, n - 1));
        }
    }

    public static class MyExtensions
    {
        public static string Reverse(this string sourse)
        {
            char[] chrAry = sourse.ToCharArray();
            Array.Reverse(chrAry);
            return new string(chrAry);
        }
    }

    public class MyInputStream
    {
        private char separator = ' ';
        private Queue<string> inputStream;
        public MyInputStream(char separator = ' ')
        {
            this.separator = separator;
            inputStream = new Queue<string>();
        }

        public string Read()
        {
            if (inputStream.Count != 0) return inputStream.Dequeue();
            string[] tmp = Console.ReadLine().Split(separator);
            for (int i = 0; i < tmp.Length; i++)
            {
                inputStream.Enqueue(tmp[i]);
            }
            return inputStream.Dequeue();
        }
        public string ReadLine() => Console.ReadLine();
        public int ReadInt() => int.Parse(Read());
        public long ReadLong() => long.Parse(Read());
        public double ReadDouble() => double.Parse(Read());
    }

    public class Pair<T1, T2>
    {
        public T1 First;
        public T2 Second;
        public Pair() { First = default(T1); Second = default(T2); }
        public Pair(T1 f, T2 s) { First = f; Second = s; }
        public override string ToString() { return "(" + First + ", " + Second + ")"; }
        public override int GetHashCode() { return First.GetHashCode() ^ Second.GetHashCode(); }
        public override bool Equals(object obj)
        {
            if (ReferenceEquals(this, obj)) return true;
            else if (obj == null) return false;
            var tmp = obj as Pair<T1, T2>;
            return (object)tmp != null && First.Equals(tmp.First) && Second.Equals(tmp.Second);
        }
    }

    #region HashMap
    public class HashMap<K, V> : Dictionary<K, V>
    {
        public HashMap() : base() { }
        public HashMap(int cap) : base(cap) { }
        public HashMap(IDictionary<K, V> dictionary) : base(dictionary) { }
        new public V this[K i]
        {
            get
            {
                V v;
                return TryGetValue(i, out v) ? v : base[i] = default(V);
            }
            set { base[i] = value; }
        }
    }
    #endregion
}