There are N integers X_1, X_2, \cdots, X_N, and we know that A_i \leq X_i \leq B_i. Find the number of different values that the median of X_1, X_2, \cdots, X_N can take.

Constraints

* 2 \leq N \leq 2 \times 10^5
* 1 \leq A_i \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_2 B_2
:
A_N B_N


Output

Print the answer.

Examples

Input

2
1 2
2 3


Output

3


Input

3
100 100
10 10000
1 1000000000


Output

9991
import java.io.*;
import java.util.*;
import static java.lang.System.out;

public class Main {
    static MyReader in = new MyReader();

    public static void main(String[] args) {
        int N = in.i();
        int[] A = new int[N];
        int[] B = new int[N];
        for (int i = 0; i < N; i++) {
            A[i] = in.i();
            B[i] = in.i();
        }
        Arrays.sort(A);
        Arrays.sort(B);
        int ans;
        if (N % 2 == 1) {
            ans = B[B.length / 2] - A[A.length / 2] + 1;
        } else {
            ans = B[B.length / 2 - 1] + B[B.length / 2] - A[A.length / 2 - 1] - A[A.length / 2] + 1;
        }
        out.println(ans);
    }
}

class MyReader extends BufferedReader {
    char[] cbuf = new char[1024];
    int head = 0;
    int tail = 0;

    MyReader() {
        super(new InputStreamReader(System.in));
    }

    char next() {
        if (head == tail) {
            try {
                tail = super.read(cbuf, 0, cbuf.length);
            } catch (IOException e) {
                e.printStackTrace();
            }
            head = 0;
        }
        return cbuf[head++];
    }

    void back() {
        head--;
    }

    boolean minus() {
        boolean minus;
        while (true) {
            char c = next();
            if (!isDelimiter(c)) {
                if (!(minus = c == '-')) back();
                return minus;
            }
        }
    }

    void skip() {
        while (isDelimiter(next()));
        back();
    }

    char[] s(int N) {
        char[] cbuf = new char[N];
        read(cbuf, 0, N);
        return cbuf;
    }

    public int read(char[] cbuf, int off, int len) {
        skip();
        int i;
        for (i = 0; i < cbuf.length; i++) {
            char c = next();
            if (isDelimiter(c)) {
                break;
            }
            cbuf[i] = c;
        }
        return i;
    }

    boolean isDelimiter(char c) {
        return c == ' ' || c == '\n' || c == '\r';
    }

    int i() {
        boolean minus = minus();
        int n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }

    int[] ii(final int N) {
        int[] a = new int[N];
        for (int j = 0; j < a.length; j++) a[j] = i();
        return a;
    }

    long l() {
        boolean minus = minus();
        long n = 0;
        while (true) {
            int k = next() - '0';
            if (k < 0 || 9 < k) break;
            n = 10 * n + k;
        }
        return minus ? -n : n;
    }
}
