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.*;
import java.io.*;
import java.nio.charset.StandardCharsets;

public class Main {
	int N;
	long[] A, B;

	public static void main(String[] args) throws IOException {
		InputStreamReader reader = new InputStreamReader(System.in, StandardCharsets.UTF_8);
		BufferedReader in = new BufferedReader(reader);
		Main ins = new Main(in);
		ins.calc();
		ins.showResult();
	}

	Main(BufferedReader in) throws IOException {
		N = Integer.parseInt(in.readLine());
		A = new long[N];
		B = new long[N];
		for (int i = 0; i < N; ++i) {
			String[] t = in.readLine().split(" ");
			A[i] = Long.parseLong(t[0]);
			B[i] = Long.parseLong(t[1]);
		}
		Arrays.sort(A);
		Arrays.sort(B);
	}

	void calc() {
		if (N % 2 == 0) {
			long aSmallMiddle = A[N / 2 - 1];
			long aLargeMiddle = A[N / 2];
			long bSmallMiddle = B[N / 2 - 1];
			long bLargeMiddle = B[N / 2];
			long result = Math.abs(bSmallMiddle - aLargeMiddle) * 2 + 1;
			if (aSmallMiddle != aLargeMiddle) {
				result += 1;
			}
			if (bSmallMiddle != bLargeMiddle) {
				result += 1;
			}
			if (bSmallMiddle < aLargeMiddle) {
				result -= 2;
			}
			System.out.println(result);
		} else {
			long aMiddle = A[N / 2];
			long bMiddle = B[N / 2];
			long result = bMiddle - aMiddle + 1;
			System.out.println(result);
		}
	}

	void showResult() {
	}
}