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 C# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

using System;
using System.IO;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder.Contest.E
{
	static class Program
	{
		public static void Main(string[] args)
		{
			var sw = new StreamWriter(Console.OpenStandardOutput()) { AutoFlush = false };
			Console.SetOut(sw);

			var cin = new Scanner();

			int n = cin.NextInt();
			var a = new int[n];
			var b = new int[n];
            for (int i = 0; i < n; i++)
            {
				a[i] = cin.NextInt();
				b[i] = cin.NextInt();
            }
			Array.Sort(a);
			Array.Sort(b);
			long ans = 0;
            if (n % 2 == 0)
            {
				double midmin = (a[(n + 1) / 2 - 1] + a[(n + 1) / 2] / 2);
				double midmax = (b[(n + 1) / 2 - 1] + b[(n + 1) / 2] / 2);
                ans = (long)((midmax - midmin) * 2) + 1;

            }
            else
            {
				long midmin = a[n / 2];
				long midmax = b[n / 2];
				ans = midmax - midmin + 1;
            }
            Console.WriteLine(ans);

			Console.Out.Flush();
		}
	}

	class Scanner
	{
		string[] s;
		int i;

		char[] cs = new char[] { ' ' };

		public Scanner()
		{
			s = new string[0];
			i = 0;
		}

		public string Next()
		{
			if (i < s.Length) return s[i++];
			string st = Console.ReadLine();
			while (st == "") st = Console.ReadLine();
			s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
			if (s.Length == 0) return Next();
			i = 0;
			return s[i++];
		}

		public int NextInt()
		{
			return int.Parse(Next());
		}

		public int[] ArrayInt(int N, int add = 0)
		{
			int[] Array = new int[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextInt() + add;
			}
			return Array;
		}

		public long NextLong()
		{
			return long.Parse(Next());
		}

		public long[] ArrayLong(int N, long add = 0)
		{
			long[] Array = new long[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextLong() + add;
			}
			return Array;
		}

		public double NextDouble()
		{
			return double.Parse(Next());
		}

		public double[] ArrayDouble(int N, double add = 0)
		{
			double[] Array = new double[N];
			for (int i = 0; i < N; i++)
			{
				Array[i] = NextDouble() + add;
			}
			return Array;
		}
	}
}