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;

namespace ABC169test
{
	class Program
	{
		static void Main(string[] args)
		{
			long N = Convert.ToInt64(Console.ReadLine());
			long[] A = new long[N];
			long[] B = new long[N];
			string[] s = new string[2];

			long MIN, MAX;
			long below = 0, above;		//基準より上下（基準含まない）
			long b = 0, a = 0;	//次に追加されるINDEX
			long count = 0;
			bool check;

			for(int i=0; i<N; i++)
			{
				s = Console.ReadLine().Split(" ");
				A[i] = Convert.ToInt64(s[0]);
				B[i] = Convert.ToInt64(s[1]);
			}
			Array.Sort(A);
			Array.Sort(B);
			MIN = A[0];
			MAX = B[N-1];

			above = Array.FindAll(A, item => item > MIN).Length;
			a = N - above;

			for (long i=MIN; i<=MAX; i++)
			{
				if (below <= N / 2 && above <= N / 2)
					count++;

				check = true;
				while (check)
				{
					check = false;
					if (b < N && B[b] == i)
					{
						below++;
						b++;
						check = true;
					}
					if (a < N && A[a] == i + 1)
					{
						above--;
						a++;
						check = true;
					}
				}

				//below = Array.FindAll(B, item => item < i).Length;
				//above = Array.FindAll(A, item => item > i).Length;

			}

			Console.WriteLine(count);
		}
	}
}