Problem:
Score: 100 points

Problem Statement

We have two distinct integers A and B.

Print the integer K such that |A - K| = |B - K|.

If such an integer does not exist, print IMPOSSIBLE instead.

Constraints

All values in input are integers.

0 \leq A,\ B \leq 10^9

A and B are distinct.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the integer K satisfying the condition.

If such an integer does not exist, print IMPOSSIBLE instead.

Sample Input 1

2 16

Sample Output 1

9

|2 - 9| = 7 and |16 - 9| = 7, so 9 satisfies the condition.

Sample Input 2

0 3

Sample Output 2

IMPOSSIBLE

No integer satisfies the condition.

Sample Input 3

998244353 99824435

Sample Output 3

549034394

Solution in C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;
using System.Linq;

namespace cswork
{
    class Program
    {
        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);
                i = 0;
                return next();
            }

            public int nextInt()
            {
                return int.Parse(next());
            }

            public long nextLong()
            {
                return long.Parse(next());
            }

            public double nextDouble()
            {
                return double.Parse(next());
            }
        }

        public Program() { }

        static void Main(string[] args)
        {
            new Program().Omame();
        }

        Scanner cin;
        const int MAXIN = 1024;
        bool[] used = new bool[MAXIN];
        int[] perm = new int[MAXIN];

        List<int[]> permList = new List<int[]>();

        public void Permutation(int pos, int n, int length)
        {
            if (pos == n)
            {
                permList.Add((int[])perm.Clone());
            }

            for (int i = 0; i < n; i++)
            {
                if (!used[i])
                {
                    perm[pos] = i;
                    used[i] = true;
                    Permutation(pos + 1, n, length);
                    used[i] = false;
                }
            }
        }

        public static List<long> GeneratePrime(long max)
        {
            long prime;
            double sqrtMax = Math.Sqrt(max);
            var primeList = new List<long>();

            List<long> searchList = new List<long>();
            for (long i = 2; i <= max - 1; i++)
            {
                searchList.Add(i);
            }

            do
            {
                prime = searchList.First();
                primeList.Add(prime);
                searchList.RemoveAll(n => n % prime == 0);
            } while (prime < sqrtMax);

            primeList.AddRange(searchList);

            return primeList;
        }

        // 最小公倍数
        public static long Lcm(long a, long b)
        {
            return a * b / Gcd(a, b);
        }

        // ユークリッドの互除法 
        public static long Gcd(long a, long b)
        {
            if (a < b)
                return Gcd(b, a);
            while (b != 0)
            {
                var remainder = a % b;
                a = b;
                b = remainder;
            }
            return a;
        }

        public static readonly string Yes = "Yes";
        public static readonly string YES = "YES";
        public static readonly string No = "No";
        public static readonly string NO = "NO";
        public static readonly string OK = "OK";
        public static readonly string NG = "NG";

        void Omame()
        {
            cin = new Scanner();
            
            long A = cin.nextLong();
            long B = cin.nextLong();

            if(A%2 == 0 && B%2 == 0)
            {
                Console.WriteLine((A + B) / 2);
            }else if (A % 2 == 1 && B % 2 == 1)
            {
                Console.WriteLine((A + B) / 2);
            }
            else
            {
                Console.WriteLine("IMPOSSIBLE");
            }
        }
    }
}