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.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.Collectors;

public class Main {

  static int n;
  static List<Long> a, b;
  static long ans;

  public static void main(String[] args) throws Exception {

    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter out = new PrintWriter(System.out);

    String[] in = br.readLine().split(" ");
    n = Integer.parseInt(in[0]);

    a = new ArrayList<>();
    b = new ArrayList<>();

    for (int i = 0; i < n; i++) {

      in = br.readLine().split(" ");
      a.add(Long.parseLong(in[0]));
      b.add(Long.parseLong(in[1]));
    }

    br.close();

    List<Long> sa = a.stream().sorted().collect(Collectors.toList());
    List<Long> sb = b.stream().sorted().collect(Collectors.toList());

    if(n % 2 == 0){
      ans = ((sb.get(n/2) + sb.get(n/2 - 1)) - (sa.get(n/2) + sa.get(n/2 - 1))) + 1;
    }else{
      ans = sb.get(n/2) -  sa.get(n / 2) + 1;
    }

    out.println(ans);

    out.flush();
  }
}
