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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.util.InputMismatchException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.Comparator;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskC {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            long[] a = new long[n];
            long[] b = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = in.nextLong();
                b[i] = in.nextLong();
            }

            List<LongLongPair> list = new ArrayList<>();
            for (int i = 0; i < n; i++) {
                list.add(new LongLongPair(a[i], b[i]));
            }
            list.sort(Comparator.comparingLong(p -> p.first + p.second));

            long ans = 0;
            for (int i = 0; i < n; i++) {
                if (i % 2 == 0) {
                    ans += list.get(n - 1 - i).first;
                } else {
                    ans -= list.get(n - 1 - i).second;
                }
            }

            out.println(ans);
        }

    }

    static class LongLongPair implements Comparable<LongLongPair> {
        public final long first;
        public final long second;

        public LongLongPair(long first, long second) {
            this.first = first;
            this.second = second;
        }

        public boolean equals(Object o) {
            if (this == o) {
                return true;
            }
            if (o == null || getClass() != o.getClass()) {
                return false;
            }

            LongLongPair pair = (LongLongPair) o;

            return first == pair.first && second == pair.second;
        }

        public int hashCode() {
            int result = Long.hashCode(first);
            result = 31 * result + Long.hashCode(second);
            return result;
        }

        public String toString() {
            return "(" + first + "," + second + ")";
        }

        public int compareTo(LongLongPair o) {
            int value = Long.compare(first, o.first);
            if (value != 0) {
                return value;
            }
            return Long.compare(second, o.second);
        }

    }

    static class InputReader {
        private BufferedReader br;
        private StringTokenizer st;

        public InputReader(InputStream inputStream) {
            br = new BufferedReader(new InputStreamReader(inputStream));
            st = new StringTokenizer("");
        }

        public String nextString() {
            while (!st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine(), " ");
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
            }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(nextString());
        }

        public long nextLong() {
            return Long.parseLong(nextString());
        }

    }
}