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: Wrong Answer)

import java.io.IOException;
import java.util.NoSuchElementException;

import java.io.InputStream;

import java.io.PrintWriter;

import java.util.Arrays;

@SuppressWarnings("unchecked")
class DishT implements Comparable<DishT> {
    int t;
    int a;
    int pos;

    DishT(int t, int a, int pos) {
        this.t = t;
        this.a = a;
        this.pos = pos;
    }

    // 高橋
    // 降順
    @Override
    public int compareTo(DishT d) {
        int ret = 0;
        if (this.t < d.t) ret = 1;
        else if (d.t < this.t) ret = -1;
        else if (this.a < d.a) ret = 1;
        else if (d.a < this.a) ret = -1;
        return ret;
    }
}

class DishA implements Comparable<DishA> {
    int t;
    int a;
    int pos;

    DishA(int t, int a, int pos) {
        this.t = t;
        this.a = a;
        this.pos = pos;
    }

    // 青木
    // 降順
    @Override
    public int compareTo(DishA d) {
        int ret = 0;
        if (this.a < d.a) ret = 1;
        else if (d.a < this.a) ret = -1;
        else if (this.t < d.t) ret = 1;
        else if (d.t < this.t) ret = -1;
        return ret;
    }
}

public class Main {
    public static void main(String[] args) throws IOException {
        PrintWriter out = new PrintWriter(System.out);

        final int N = nextInt();

        DishT[] T = new DishT[N];
        DishA[] A = new DishA[N];

        for (int i = 0; i < N; i++) {
            int t = nextInt();
            int a = nextInt();
            T[i] = new DishT(t, a, i);
            A[i] = new DishA(t, a, i);
        }

        Arrays.sort(T);
        Arrays.sort(A);

        boolean done[] = new boolean[N];

        int cnt = 0;
        int tidx = 0;
        int aidx = 0;
        long ans = 0;

        while (cnt < N) {
            boolean flg = true;
            while (flg) {
                if (N <= cnt) break;

                if (!done[T[tidx].pos]) {
                    ans += T[tidx].t;
                    done[T[tidx].pos] = true;
                    flg = false;
                    cnt++;
                }

                tidx++;
            }

            flg = true;
            while (flg) {
                if (N <= cnt) break;

                if (!done[A[aidx].pos]) {
                    ans -= T[aidx].a;
                    done[A[aidx].pos] = true;
                    flg = false;
                    cnt++;
                }

                aidx++;
            }
        }

        out.println(ans);

        out.flush();
    }

    // FastScanner start
    static final InputStream in = System.in;
    static final byte[] buffer = new byte[1024];
    static int ptr = 0;
    static int buflen = 0;
    static boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        } else {
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    static int readByte() {
        if (hasNextByte()) return buffer[ptr++];
        else return -1;
    }
    static boolean isPrintableChar(int c) {
        return 33 <= c && c <= 126;
    }
    static boolean hasNext() {
        while (hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++;
        return hasNextByte();
    }
    static String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while (isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    static long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while (true) {
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            } else if(b == -1 || !isPrintableChar(b)) {
                return minus ? -n : n;
            } else {
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    static int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
            throw new NumberFormatException();
        return (int) nl;
    }
    static double nextDouble() {
        return Double.parseDouble(next());
    }
    // FastScanner end
}