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;
using System.Threading.Tasks;
using System.Xml.Schema;

namespace E
{
    class Program
    {
        static void Main(string[] args)
        {
            string line = Console.ReadLine();
            int N = int.Parse(line);
            List<int> A = new List<int>();
            List<int> B = new List<int>();
            for(int i = 0; i < N; i++)
            {
                line = Console.ReadLine();
                string[] strs = line.Split(' ');
                A.Add(int.Parse(strs[0]));
                B.Add(int.Parse(strs[1]));
            }
            A.Sort();
            B = B.OrderByDescending(x => x).ToList();
            if (N % 2 == 0)
            {
                // Nが偶数 -> 中央が定まらない -> 2分の1単位で数える
                // N / 2 と N / 2 + 1の間？
                double MAX = (B[N / 2 - 1] + B[N / 2]);
                double MIN = (A[N / 2 - 1] + A[N / 2]);
                Console.WriteLine((MAX - MIN + 1));
            }
            else
            {
                // Nが奇数 -> 中央が定まる
                int MAX = B[N / 2];
                int MIN = A[N / 2];
                Console.WriteLine(MAX - MIN + 1);
            }
        }
    }
}