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.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] A = new int[n], B = new int[n];
        int[][] sA = new int[n][2], sB = new int[n][2];
        for (int i = 0; i < n; i++) {
            A[i] = sc.nextInt();
            B[i] = sc.nextInt();
            sA[i][0] = A[i]; sA[i][1] = i;
            sB[i][0] = B[i]; sB[i][1] = i;
        }
        Arrays.sort(sA, (x,y) -> x[0] - y[0]);
        Arrays.sort(sB, (x,y) -> y[0] - x[0]);
        int ans = 0;
        int minMin = sA[(n-1) / 2][0];
        int maxMax = sB[(n-1) / 2][0];
        if(n % 2 == 1){
            ans = maxMax - minMin +1;
        }else{
            int minMax = 0;
            int maxMin = Integer.MAX_VALUE;
            for (int i = 0; i <= (n-1)/2; i++) {
                if(minMax < B[sA[i][1]]) minMax = B[sA[i][1]];
                if(A[sB[i][1]] < maxMin) maxMin = A[sB[i][1]];
            }
            minMax = Math.min(minMax, maxMax);
            maxMin = Math.max(maxMin, minMin);
            ans = minMax - minMin + maxMax - maxMin + 1;
        }
        System.out.println(ans);
        sc.close();

    }

}