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.util.Arrays;
import java.util.Scanner;

public class Main {
	static long median(long array[]){
		int N = array.length;
		if(N % 2 == 0){
			return (array[N / 2] + array[N / 2 - 1]) ;			
		}else{
			return array[N / 2];
		}
	}
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		long A[] = new long[N];
		long 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);
		if(N % 2 == 1){
			long ret = median(B) - median(A) + 1;
			System.out.println(ret);			
		}else{
			long ret = median(B) - median(A) + 1;
			System.out.println(ret);			
		}
	}
}