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: Wrong Answer)

import math

N = int(input())
A = []
B = []
for i in range(N):
    in1, in2 = list(map(int,input().split()))
    A.append(in1)
    B.append(in2)
AA = sorted(A)
BB = sorted(B)

if N%2==1:
    low = AA[N//2  ]
    high = BB[N//2  ]
    print(high - low +1)

else:
    low = (AA[N//2-1] + AA[N//2]) / 2
    high = (BB[N//2-1] + BB[N//2]) / 2
    count=0
    while True:
        low = low+ 0.5*count
        count+=1
        if low>high:
            break
    #low = math.floor(low)
    #high = math.ceil(high)
    print(count)