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: Accepted)

using System;
using System.Collections.Generic;

namespace ConsoleApp1
{

    class Program
    {

        static void Main(string[] args)
        {
            int n = int.Parse(Console.ReadLine());
            var a = new List<int> { };
            var b = new List<int> { };

            for (int i = 0; i < n; i++)
            {
                var input = Console.ReadLine().Split(' ');
                a.Add(int.Parse(input[0]));
                b.Add(int.Parse(input[1]));
            }

            a.Sort();
            b.Sort();
            int midA = 0;
            int midB = 0;
            int count = 0;
            if (n % 2 == 0)
            {
                midA = a[n / 2 - 1] + a[n / 2];
                midB = b[n / 2 - 1] + b[n / 2];
                count = midB - midA + 1;
            }
            else
            {
                midA = a[(n - 1) / 2];
                midB = b[(n - 1) / 2];
                count = midB - midA + 1;
            }

            Console.WriteLine(count);

        }

    }
}