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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
	
	public static void main(String[] args) throws IOException {
		MyScanner sc = new MyScanner();
		PrintWriter out = new PrintWriter(System.out);
		int N = sc.nextInt();
		int[] A = new int[N], B = new int[N];
		for (int i = 0; i < N; i++) {
			A[i] = sc.nextInt();
			B[i] = sc.nextInt();
		}
		Arrays.sort(A);
		Arrays.sort(B);
		boolean impar = N % 2 == 1;
		int ans = 0;
		if (impar) {
			int min = A[N/2];
			int max = B[N/2];
			ans = max - min + 1;
		} else {
			int min = A[N/2-1] + A[N/2];
			int max = B[N/2-1] + B[N/2];
			ans = max - min + 1;
		}
		out.println(ans);
		
		out.flush();
	}
	
	static class MyScanner {
		private BufferedReader br;
		private StringTokenizer tokenizer;
		
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}
