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 PyPy3 (7.3.0) (judge verdict: Accepted)

import math
from functools import reduce
from collections import deque
import sys
sys.setrecursionlimit(10**7)

from statistics import median

def input():
    return sys.stdin.readline().strip()

# スペース区切りの入力を読み込んで数値リストにして返します。
def get_nums_l():
    return [ int(s) for s in input().split(" ")]

# 改行区切りの入力をn行読み込んで数値リストにして返します。
def get_nums_n(n):
    return [ int(input()) for _ in range(n)]

# 改行またはスペース区切りの入力をすべて読み込んでイテレータを返します。
def get_all_int():
    return map(int, open(0).read().split())

def rangeI(it, l, r):
    for i, e in enumerate(it):
        if l <= i < r:
            yield e
        elif l >= r:
            break

def log(*args):
    print("DEBUG:", *args, file=sys.stderr)

INF = 999999999999999999999999
MOD = 10**9+7

n = int(input())
AB = []
for _ in range(n):
    AB.append(get_nums_l())
A = [ ab[0] for ab in AB ]
B = [ ab[1] for ab in AB ]

if n%2 == 1:
    min_of_med = median(A)
    max_of_med = median(B)
    print(max_of_med - min_of_med + 1)

else:
    A.sort()
    B.sort()
    min_of_med = A[n//2 - 1] + A[n//2]
    max_of_med = B[n//2 - 1] + B[n//2]
    print(max_of_med - min_of_med + 1)