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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();

		long[] a = new long[n], b = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextLong();
			b[i] = sc.nextLong();
		}
		Arrays.sort(a);
		Arrays.sort(b);

		int mid = n / 2;
		if (n % 2 == 0) { // .5も含む
//			double min = (a[mid - 1] + a[mid]) / 2;
//			double max = (b[mid - 1] + b[mid]) / 2;
//			System.out.println((int) (max - min) * 2 + 1); // (max-min+1)*2-1
			
			// 整数で扱っても結果は変わらない．
			long min = (a[mid - 1] + a[mid]) / 2;
			long max = (b[mid - 1] + b[mid]) / 2;
			System.out.println((max - min) * 2 + 1);
		} else {
			// 整数値のみ
			System.out.println(b[mid] - a[mid] + 1);
		}
	}
}