Problem:
Score : 300 points

Problem Statement

Takahashi, Nakahashi and Hikuhashi have integers A, B and C, respectively.
After repeating the following operation K times, find the integer Takahashi will get minus the integer Nakahashi will get:

Each of them simultaneously calculate the sum of the integers that the other two people have, then replace his own integer with the result.

However, if the absolute value of the answer exceeds 10^{18}, print Unfair instead.

Constraints

1 \leq A,B,C \leq 10^9

0 \leq K \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C K

Output

Print the integer Takahashi will get minus the integer Nakahashi will get, after repeating the following operation K times.
If the absolute value of the answer exceeds 10^{18}, print Unfair instead.

Sample Input 1

1 2 3 1

Sample Output 1

1

After one operation, Takahashi, Nakahashi and Hikuhashi have 5, 4 and 3, respectively. We should print 5-4=1.

Sample Input 2

2 3 2 0

Sample Output 2

-1

Sample Input 3

1000000000 1000000000 1000000000 1000000000000000000

Sample Output 3

0

Solution in C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

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

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            Scanner cin = new Scanner();
            long A = cin.Long();
            long B = cin.Long();
            long C = cin.Long();
            long K = cin.Long();

            if (K % 2 == 0) Console.WriteLine(B - A);
            else Console.WriteLine(A - B);

            Console.ReadLine();
        }

        static int GCD(int a, int b) //a,bの最大公約数を返す
        {
            if (a < b)
                // 引数を入替えて自分を呼び出す
                return GCD(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }

        static int LCM(int a, int b) //a,bの最小公倍数を返す
        {
            return a / GCD(a, b) * b;
        }
    

        static int BinarySearch<T>(T[] array, T target) where T : IComparable<T> //2分探索
        {
            int min = 0;
            int max = array.Length - 1;

            while (true)
            {
                // 範囲内に見つからなかった
                if (max < min) return -1;
                // 範囲内の中央値と比較する
                int mid = min + (max - min) / 2;
                switch (target.CompareTo(array[mid]))
                {
                    case -1:    // 中央値より小さい
                        max = mid - 1;
                        break;
                    case 1:     // 中央値より大きい
                        min = mid + 1;
                        break;
                    case 0:     // 見つかった
                        return mid;
                }
            }
        }

        static void Swap<T>(ref T a, ref T b)
        {
            T tmp = a;
            a = b;
            b = tmp;
        }
    }

    class Scanner //標準入力
    {
        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;
        }
    }
}