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 Python (3.8.2) (judge verdict: Accepted)


import glob

# 問題ごとのディレクトリのトップからの相対パス
REL_PATH = 'ABC\\169\\E'

# テスト用ファイル置き場のトップ
TOP_PATH = 'C:\\AtCoder'


class Common:
    problem = []
    index = 0

    def __init__(self, rel_path):
        self.rel_path = rel_path

    def initialize(self, path):
        file = open(path)
        self.problem = file.readlines()
        self.index = 0
        return

    def input_data(self):
        try:
            IS_TEST
            self.index += 1
            return self.problem[self.index - 1]

        except NameError:
            return input()

    def resolve(self):
        pass

    def exec_resolve(self):
        try:
            IS_TEST
            for path in glob.glob(TOP_PATH + '\\' + self.rel_path + '/*.txt'):
                print("Test: " + path)
                self.initialize(path)
                self.resolve()
                print("\n\n")
        except NameError:
            self.resolve()


class Solver(Common):

    def resolve(self):

        N = int(self.input_data())
        A = [[0, 0] for i in range(N)]
        B = [[0, 0] for i in range(N)]

        for i in range(N):
            tmp = self.input_data().split()
            A[i][0] = i
            A[i][1] = int(tmp[0])
            B[i][0] = i
            B[i][1] = int(tmp[1])

        # 最小値群と最大値群を個別に並び替える
        A_sorted = sorted(A, key=lambda x: x[1])
        B_sorted = sorted(B, key=lambda x: x[1])
        '''
        if N % 2 == 0:
            for i in range(N):
                for j in range(N):
                    if A_sorted[j][0] == i:
                        min = A_sorted[j]
                        break
                for j in range(N):
                    if B_sorted[j][0] == i:
                        max = B_sorted[j]
                        break
                

            pass
        else:
            pass
        '''
        if N % 2 == 0:
            median_min = A_sorted[int(N/2)][1] + A_sorted[int(N/2)-1][1]
            median_max = B_sorted[int(N/2)][1] + B_sorted[int(N/2)-1][1]
        else:
            median_min = A_sorted[int(N/2)][1]
            median_max = B_sorted[int(N/2)][1]
        result = median_max - median_min + 1

        print(str(result))



solver = Solver(REL_PATH)
solver.exec_resolve()