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: Wrong Answer)

import java.math.BigDecimal;
import java.util.*;

public class Main {

    Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Main().run();
    }

    void run() {

        int n = sc.nextInt();
        int[][] arr = new int[n][2];

        Data[] data = new Data[n];
        for (int i=0; i<n; i++) {
            arr[i][0] = sc.nextInt();
            arr[i][1] = sc.nextInt();
            data[i] = new Data(arr[i][0], arr[i][1]);
        }

        Arrays.sort(data);

//        System.out.println(data[0].b);

        if (n%2 == 1) {
            System.out.println(data[n%2].b-data[n%2].a+1);
            return;
        }

        int min = data[n/2-1].a + data[n/2].a;


        Data2[] data2s = new Data2[n];
        for (int i=0; i<n; i++) {
            data2s[i] = new Data2(arr[i][0], arr[i][1]);
        }

        Arrays.sort(data2s);
        int max = data2s[n/2-1].b + data2s[n/2].b;

        System.out.println((max-min+1));
    }
    public static class Data implements Comparable<Data> {
        int a, b;
        public Data(int a, int b) {
            this.a = a;
            this.b = b;
        }
        public int compareTo(Data other) {
            if (this.a == other.a) {
                return this.b - other.b;
            } else {
                return this.a - other.a;
            }
        }
    }
    public static class Data2 implements Comparable<Data2> {
        int a, b;
        public Data2(int a, int b) {
            this.a = a;
            this.b = b;
        }
        public int compareTo(Data2 other) {
            if (this.a == other.a) {
                return this.b - other.b;
            } else {
                return other.a - this.a;
            }
        }
    }
}