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.

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.

Examples

Input

2
1 2
2 3


Output

3


Input

3
100 100
10 10000
1 1000000000


Output

9991
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] * 2;
			long aLargeMiddle = A[N / 2] * 2;
			long bSmallMiddle = B[N / 2 - 1] * 2;
			long bLargeMiddle = B[N / 2] * 2;
			long small = (aLargeMiddle + aSmallMiddle) / 2;
			long large = (bLargeMiddle + bSmallMiddle) / 2;
			long result = large - small + 1;
			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() {
	}
}