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.Collections.Generic;
using System.Linq;

namespace ProblemE
{
    class Program
    {
        class Range
        {
            public int Max { get; set; } 
            public int Min { get; set; }

            public Range(int max, int min)
            {
                Max = max;
                Min = min;
            }
        }

        static void Main()
        {
            int N = int.Parse(Console.ReadLine());
            var A = new List<int>();
            var B = new List<int>();
            var range = new Range[N];
            for(int i = 0; i < N; i++)
            {
                var temp = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
                range[i] = new Range(temp[1], temp[0]);
                A.Add(temp[0]);
                B.Add(temp[1]);
            }
            A.Sort();
            B.Sort((a, b) => b - a);

            
            var ans = new List<int>();
            if(N % 2 == 1)
            {                
                int min = A[N / 2];
                int max = B[N / 2];

                Console.WriteLine(max - min + 1);
            }
            else
            {
                var listx = new List<int>();
                var listy = new List<int>();

                int min = A[N / 2 - 1];
                
                int max = B[N / 2 - 1];

                Console.WriteLine((max - min) * 2 - 1);
            }

           
        }
    }
}