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 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main {
 
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        final int N = sc.nextInt();
        List<Integer> min = new ArrayList<>();
        List<Integer> max = new ArrayList<>();
        int[][] ab = new int[N][2];
        for(int i=0;i<N;i++){
            min.add(sc.nextInt());
            max.add(sc.nextInt());
        }
        Collections.sort(min);
        Collections.sort(max);
        if(N%2 == 1){
            System.out.println(max.get(N/2+1) - min.get(N/2+1) + 1);
        }else{
            System.out.println((max.get(N/2) + max.get(N/2-1) - min.get(N/2) + min.get(N/2-1)) / 2 + 1);
        }
    }
}