Problem:
Score : 500 points

Problem Statement

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.

Notes

The median of X_1, X_2, \cdots, X_N is defined as follows. Let x_1, x_2, \cdots, x_N be the result of sorting X_1, X_2, \cdots, X_N in ascending order.

If N is odd, the median is x_{(N+1)/2};

if N is even, the median is (x_{N/2} + x_{N/2+1}) / 2.

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.

Sample Input 1

2
1 2
2 3

Sample Output 1

3

If X_1 = 1 and X_2 = 2, the median is \frac{3}{2};

if X_1 = 1 and X_2 = 3, the median is 2;

if X_1 = 2 and X_2 = 2, the median is 2;

if X_1 = 2 and X_2 = 3, the median is \frac{5}{2}.

Thus, the median can take three values: \frac{3}{2}, 2, and \frac{5}{2}.

Sample Input 2

3
100 100
10 10000
1 1000000000

Sample Output 2

9991

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.io.InputStream;
import java.io.PrintStream;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

  public static void main(String[] args) {
    resolve(System.in, System.out);
  }

  static void resolve(InputStream is, PrintStream pw) {
    Scanner sc = new Scanner(is);
    int n = Integer.parseInt(sc.next());
    int min[] = new int[n];
    int max[] = new int[n];
    for (int i = 0; i < n; i++) {
      min[i] = Integer.parseInt(sc.next());
      max[i] = Integer.parseInt(sc.next());
    }
    Arrays.sort(min);
    Arrays.sort(max);
    if (n % 2 == 1) {
      pw.println(
        (int) median(max) - (int) median(min) + 1
      );
    } else {
      pw.println((int) ((median(max) - median(min)) * 2) + 1);
    }
  }

  static double median(int sorted[]) {
    if (sorted.length % 2 == 1) {
      return sorted[sorted.length / 2];
    }
    return (sorted[sorted.length / 2 - 1] + sorted[sorted.length / 2]) / 2d;
  }
}