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;
using System.Linq;
using System.Text;

namespace AtCoder169
{
    public class ProgramE
    {
        static void Main(string[] args)
        {
            var inputNumStr = Console.ReadLine();
            var inputNum = int.Parse(inputNumStr);

            var aList = new List<long>();
            var bList = new List<long>();

            for (var i = 0; i < inputNum; i++)
            {
                var input = Console.ReadLine();
                var vallist = input.Split(" ").Select(long.Parse);

                var ai = vallist.First();
                var bi = vallist.Skip(1).First();

                aList.Add(ai);
                bList.Add(bi);
            }

            if (inputNum % 2 == 1)
            {
                var a = aList.OrderBy(d => d).Skip((inputNum - 1) / 2).First();
                var b = bList.OrderBy(d => d).Skip((inputNum - 1) / 2).First();

                Console.WriteLine(b - a + 1);
            }
            else
            {
                var a1 = aList.OrderBy(d => d).Skip(inputNum / 2 - 1).First();
                var b1 = bList.OrderBy(d => d).Skip(inputNum / 2 - 1).First();
                var a2 = aList.OrderBy(d => d).Skip(inputNum / 2).First();
                var b2 = bList.OrderBy(d => d).Skip(inputNum / 2).First();

                Console.WriteLine((b1 + b2) - (a1 + a2) + 1);
            }

        }
    }
}