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 static java.lang.Math.*;

public class Main {
    public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int[][] ab = new int[n][2];
		for(int i = 0; i < n; i++){
		    ab[i][0] = sc.nextInt();
		    ab[i][1] = sc.nextInt();
		}
		
		// 計算
		int result = 0;
		int from = 0;
		int to = 0;
		if(n%2 == 1){
		    Arrays.sort(ab, (s, t) -> Integer.compare(s[0], t[0]));
		    from = ab[n/2][0];
		    Arrays.sort(ab, (s, t) -> Integer.compare(s[1], t[1]));
		    to = ab[n/2][1];
		}else{
		    Arrays.sort(ab, (s, t) -> Integer.compare(s[0], t[0]));
		    from = ab[n/2-1][0] + ab[n/2][0];
		    Arrays.sort(ab, (s, t) -> Integer.compare(s[1], t[1]));
		    to = ab[n/2-1][1] + ab[n/2][1] ;
		}
		result = to-from+1;
		
		// 出力
		System.out.println(result);
	}

	public static class Scanner {
		private BufferedReader br; private StringTokenizer tok;
		public Scanner(InputStream is) throws IOException { br = new BufferedReader(new InputStreamReader(is));}
		private void getLine() throws IOException { while(!hasNext()) tok = new StringTokenizer(br.readLine());}
		private boolean hasNext() { return tok != null && tok.hasMoreTokens();}
		public String next() throws IOException { getLine(); return tok.nextToken();}
		public int nextInt() throws IOException { return Integer.parseInt(next());}
		public long nextLong() throws IOException { return Long.parseLong(next());}
		public double nextDouble() throws IOException { return Double.parseDouble(next());}
	}
}
