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.*;

public class Main {
    public static final long mod = (long)1e9+7;
    public static final long INF = Long.MAX_VALUE/2;
    public static final int inf = Integer.MAX_VALUE/2;

    static void solve(InputReader in, PrintWriter out){
        int n = in.ni();
        long[] a = new long[n], b = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nl()*2;
            b[i] = in.nl()*2;
        }
        Arrays.sort(a);
        Arrays.sort(b);
        long l, r;
        if(n%2==0){
            l = (a[n/2-1]+a[n/2])/2;
            r = (b[n/2-1]+b[n/2])/2;

        }else{
            l = a[n/2]/2;
            r = b[n/2]/2;
        }
        out.println(r-l+1);
    }

    public static void main(String[] args){
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        solve(in, out);
        out.close();
    }
    public static class InputReader{
        private BufferedReader br;
        private StringTokenizer st;
        public InputReader(InputStream is){
            br = new BufferedReader(new InputStreamReader(is));
            st = null;
        }
        public String ns(){
            if(st == null || !st.hasMoreTokens()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (Exception e){
                    throw new RuntimeException(e);
                }
            }
            return st.nextToken();
        }
        public int ni(){
            return Integer.parseInt(ns());
        }
        public long nl(){
            return Long.parseLong(ns());
        }
        public double nd(){
            return Double.parseDouble(ns());
        }
        public char nc(){
            return ns().toCharArray()[0];
        }
        public int[] ni(int n){
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }
        public long[] nl(int n){
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = nl();
            }
            return a;
        }
        public double[] nd(int n){
            double[] a = new double[n];
            for (int i = 0; i < n; i++) {
                a[i] = nd();
            }
            return a;
        }
    }
}