Problem:
Score : 400 points

Problem Statement

There are N dishes of cuisine placed in front of Takahashi and Aoki.
For convenience, we call these dishes Dish 1, Dish 2, ..., Dish N.

When Takahashi eats Dish i, he earns A_i points of happiness; when Aoki eats Dish i, she earns B_i points of happiness.

Starting from Takahashi, they alternately choose one dish and eat it, until there is no more dish to eat.
Here, both of them choose dishes so that the following value is maximized: "the sum of the happiness he/she will earn in the end" minus "the sum of the happiness the other person will earn in the end".

Find the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".

Constraints

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 B_1
:
A_N B_N

Output

Print the value: "the sum of the happiness Takahashi earns in the end" minus "the sum of the happiness Aoki earns in the end".

Sample Input 1

3
10 10
20 20
30 30

Sample Output 1

20

In this sample, both of them earns 10 points of happiness by eating Dish 1, 20 points by eating Dish 2, and 30 points by eating Dish 3.

In this case, since Takahashi and Aoki have the same "taste", each time they will choose the dish with which they can earn the greatest happiness. Thus, first Takahashi will choose Dish 3, then Aoki will choose Dish 2, and finally Takahashi will choose Dish 1, so the answer is (30 + 10) - 20 = 20.

Sample Input 2

3
20 10
20 20
20 30

Sample Output 2

20

In this sample, Takahashi earns 20 points of happiness by eating any one of the dishes 1, 2 and 3, but Aoki earns 10 points of happiness by eating Dish 1, 20 points by eating Dish 2, and 30 points by eating Dish 3.

In this case, since only Aoki has likes and dislikes, each time they will choose the dish with which Aoki can earn the greatest happiness. Thus, first Takahashi will choose Dish 3, then Aoki will choose Dish 2, and finally Takahashi will choose Dish 1, so the answer is (20 + 20) - 20 = 20.

Sample Input 3

6
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000
1 1000000000

Sample Output 3

-2999999997

Note that the answer may not fit into a 32-bit integer.

Solution in C# (Mono 4.6.2.0) (judge verdict: Runtime Error)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Runtime.CompilerServices;
using System.Runtime.Remoting.Messaging;
using System.Text;
using System.Threading.Tasks;
namespace NikkeiPGKing
{
    class Program
    {
        static void Main(string[] args)
        {
            Problem3.Problem3Read();
        }


    }
    class Problem3
    {
        static public void Problem3Read()
        {
            int n = int.Parse(Console.ReadLine());
            List<Tuple<long, long>> dishes = new List<Tuple<long, long>>();
            for (int i = 0; i < n; i++)
            {
                string[] pair = Console.ReadLine().Split(' ');
                dishes.Add(new Tuple<long, long>(long.Parse(pair[0]), long.Parse(pair[1])));
                
            }
            Console.WriteLine(Problem3Solve(dishes));
        }
        private static int NormalNegative(List<Tuple<long, long>> dishes)
        {
            NormalSort(dishes);
            long maxDiffs = Math.Abs(dishes[0].Item1 - dishes[1].Item2);
            int maxIdx = 0;
            long maxValues = dishes[0].Item1;
            for (int i = 1; i < dishes.Count; i++)
            {
                if (maxDiffs != Math.Abs(dishes[i].Item1 - dishes[i].Item2))
                {
                    break;
                }

                if (dishes[i].Item1 > maxValues)
                {
                    maxIdx = i;
                }
            }

            return maxIdx;

        }

        private static int SpecialNegatives(List<Tuple<long, long>> dishes)
        {
            long max = long.MinValue;
            foreach (var d in dishes)
            {
                if (max < d.Item1)
                {
                    max = d.Item1;
                }
            }

            //max求めた
            int maxIdx = -1;
            long maxHappyDiff = long.MinValue;
            for (int i = 0; i < dishes.Count; i++)
            {
                if (dishes[i].Item1 == max)
                {
                    if (maxHappyDiff < Math.Abs(dishes[i].Item1 - dishes[i].Item2))
                    {
                        maxHappyDiff = Math.Abs(dishes[i].Item1 - dishes[i].Item2);
                        maxIdx = i;
                    }
                }
            }

            return maxIdx;
        }

        static public long Problem3Solve(List<Tuple<long, long>> dishes)
        {
            if (dishes.Count % 2 == 1)
            {
                var dishes1 = new List<Tuple<long, long>>(dishes);
                var dishes2 = new List<Tuple<long, long>>(dishes);
                long max1 = 0;
                int idx1 = SpecialNegatives(dishes1);
                max1 += dishes1[idx1].Item1;
                dishes1.RemoveAt(idx1);
                max1 += CalcLeft(dishes1);
                long max2 = 0;
                int idx2 = NormalNegative(dishes2);
                max2 += dishes2[idx2].Item1;
                dishes2.RemoveAt(idx2);
                max2 += CalcLeft(dishes1);
                if (max1 > max2)
                {
                    return max1;
                }
                else
                {
                    return max2;
                }
            }
            else
            {

                return CalcLeft(dishes);
            }
            
        }

        private static long CalcLeft(List<Tuple<long, long>> dishes)
        {
            long a = 0;
            long b = 0;
            NormalSort(dishes);
            //食い合いをする
            for (int i = 0; i < dishes.Count; i++)
            {
                if (i % 2 == 0)
                {
                    a += dishes[i].Item1;
                }
                else
                {
                    b += dishes[i].Item2;
                }
            }
            return a - b;
        }

        private static void NormalSort(List<Tuple<long, long>> dishes)
        {
            dishes.Sort((t1, t2) =>
            {
                long res = Math.Abs(t1.Item1 - t1.Item2) - Math.Abs(t2.Item1 - t2.Item2);
                if (res > 0)
                {
                    return -1;
                }

                if (res == 0)
                {
                    return 0;
                }

                return 1;
            }
            );
        }
    }

}