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.io.*;
import java.math.*;
import java.util.*;
public class Main { 

	public static void main(String[] args) { //
		Scanner input = new Scanner(System.in);
		int N = input.nextInt();
		long ans = 0;
		long[][] ranges = new long[N][2];
		long[] lower = new long[N];
		long[] upper = new long[N];
		for (int i = 0; i < N; i++) {
			long A = input.nextLong();
			long B = input.nextLong();
			ranges[i][0]=A;
			ranges[i][1]=B;
			lower[i]=A;
			upper[i]=B;
		}
		Arrays.sort(lower);
		Arrays.sort(upper);
		if (N%2==0) {
			int L = N/2-1;
			int R = N/2;
			long ways = upper[R]-lower[L]+1;
			ans=Math.max(0, ways);
		}else {
			int index = N/2;
			ans=upper[index]-lower[index]+1;
			ans=Math.max(0, ans);
		}
		System.out.println(ans);
	}	
}