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;
using System.Collections.Generic;
using System.Globalization;
using System.Linq;
using System.Numerics;
using System.Text;
using System.Threading;
using System.Threading.Tasks;
using static System.Console;

namespace ConsoleApp2
{
    class Program
    {
        static public int[] Sarray() { return ReadLine().Split().Select(int.Parse).ToArray(); }
        static public List<int> Slist() { return new List<int>(ReadLine().Split().Select(int.Parse)); }
        static void Main(string[] args)
        {
            var n = Sarray()[0];
            var ls = new List<int>();
            var rs = new List<int>();
            for(var i=0;i<n;i++)
            {
                var tmp = Sarray();
                ls.Add(tmp[0]);
                rs.Add(tmp[1]);
            }
            ls.Sort();
            rs.Sort();
            var ans = 0;
            if(n%2==1)
            {
                var l = ls[n / 2];
                var r = rs[n / 2];
                ans = r - l + 1;
            }
            else
            {
                var l2 = ls[n / 2 - 1] + ls[n / 2];
                var r2 = rs[n / 2 - 1] + rs[n / 2];
                ans = r2 - l2 + 1;
            }
            WriteLine(ans);
        }

    }
}