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 sys
import math
from decimal import Decimal, ROUND_HALF_UP, ROUND_HALF_EVEN
from collections import deque
from bisect import bisect_left
from itertools import product
def I(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LI2(N): return [list(map(int, sys.stdin.readline().split())) for i in range(N)]
#文字列を一文字ずつ数字に変換、'5678'を[5,6,7,8]とできる
def LSI(): return list(map(int, list(sys.stdin.readline().rstrip())))
def LSI2(N): return [list(map(int, list(sys.stdin.readline().rstrip()))) for i in range(N)]
#文字列として取得
def ST(): return sys.stdin.readline().rstrip()
def LST(): return sys.stdin.readline().rstrip().split()
def LST2(N): return [sys.stdin.readline().rstrip().split() for i in range(N)]
def FILL(i,h): return [i for j in range(h)]
def FILL2(i,h,w): return [FILL(i,w) for j in range(h)]
def FILL3(i,h,w,d): return [FILL2(i,w,d) for j in range(h)]
def FILL4(i,h,w,d,d2): return [FILL3(i,w,d,d2) for j in range(h)]
def sisha(num,digit): return Decimal(str(num)).quantize(Decimal(digit),rounding=ROUND_HALF_UP)
#'0.01'や'1E1'などで指定、整数に戻すならintをかます
MOD = 1000000007
INF = float("inf")
sys.setrecursionlimit(10**6+10)

N = I()
L = LI2(N)
A = [i[0] for i in L]
B = [i[1] for i in L]
A.sort()
B.sort()
if N%2==0:
    medA = (A[N//2-1]+A[N//2])
    medB = (B[N//2-1]+B[N//2])
    print((medB-medA)+1)
else:
    medA = A[(N-1)//2]
    medB = B[(N-1)//2]
    print(medB-medA+1)