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: Accepted)

using System;
using System.Linq;
using System.Collections.Generic;
using Debug = System.Diagnostics.Debug;
using SB = System.Text.StringBuilder;
using System.Numerics;
using static System.Math;
using Number = System.UInt32;
namespace Program {
    public class Solver {
        Random rnd = new Random(0);
        public void Solve() {
            var n = ri;
            var A = new long[n];
            var B = new long[n];
            for (int i = 0; i < n; i++) {
                A[i] = rl;
                B[i] = rl;
            }
            var id = Enumerate(n, x => x).OrderByDescending(x => A[x] + B[x]).ToArray();

            var u = 0L;
            var v = 0L;
            for (int i = 0; i < n; i++) {
                if (i % 2 == 0)
                    u += A[id[i]];
                else v += B[id[i]];
            }
            Console.WriteLine(u - v);

        }
        const long INF = 1L << 60;
        int ri { get { return sc.Integer(); } }
        long rl { get { return sc.Long(); } }
        double rd { get { return sc.Double(); } }
        string rs { get { return sc.Scan(); } }
        public IO.StreamScanner sc = new IO.StreamScanner(Console.OpenStandardInput());

        static T[] Enumerate<T>(int n, Func<int, T> f) {
            var a = new T[n];
            for (int i = 0; i < n; ++i) a[i] = f(i);
            return a;
        }
        static public void Swap<T>(ref T a, ref T b) { var tmp = a; a = b; b = tmp; }
    }
}

#region main
static class Ex {
    static public string AsString(this IEnumerable<char> ie) { return new string(ie.ToArray()); }
    static public string AsJoinedString<T>(this IEnumerable<T> ie, string st = " ") {
        return string.Join(st, ie);
    }
    static public void Main() {
        Console.SetOut(new Program.IO.Printer(Console.OpenStandardOutput()) { AutoFlush = false });
        var solver = new Program.Solver();
        solver.Solve();
        Console.Out.Flush();
    }
}
#endregion
#region Ex
namespace Program.IO {
    using System.IO;
    using System.Text;
    using System.Globalization;

    public class Printer: StreamWriter {
        public override IFormatProvider FormatProvider { get { return CultureInfo.InvariantCulture; } }
        public Printer(Stream stream) : base(stream, new UTF8Encoding(false, true)) { }
    }

    public class StreamScanner {
        public StreamScanner(Stream stream) { str = stream; }

        public readonly Stream str;
        private readonly byte[] buf = new byte[1024];
        private int len, ptr;
        public bool isEof = false;
        public bool IsEndOfStream { get { return isEof; } }

        private byte read() {
            if (isEof) return 0;
            if (ptr >= len) {
                ptr = 0;
                if ((len = str.Read(buf, 0, 1024)) <= 0) {
                    isEof = true;
                    return 0;
                }
            }
            return buf[ptr++];
        }

        public char Char() {
            byte b = 0;
            do b = read(); while ((b < 33 || 126 < b) && !isEof);
            return (char)b;
        }
        public string Scan() {
            var sb = new StringBuilder();
            for (var b = Char(); b >= 33 && b <= 126; b = (char)read()) sb.Append(b);
            return sb.ToString();
        }
        public string ScanLine() {
            var sb = new StringBuilder();
            for (var b = Char(); b != '\n' && b != 0; b = (char)read()) if (b != '\r') sb.Append(b);
            return sb.ToString();
        }
        public long Long() { return isEof ? long.MinValue : long.Parse(Scan()); }
        public int Integer() { return isEof ? int.MinValue : int.Parse(Scan()); }
        public double Double() { return isEof ? double.NaN : double.Parse(Scan(), CultureInfo.InvariantCulture); }
    }
}

#endregion