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 sys
import math

N = int(sys.stdin.readline().strip())
ABlist = []
for _ in range(N):
    a, b = map(int, sys.stdin.readline().strip().split())
    ABlist.append((a, b))

Asorted = sorted(ABlist)
Bsorted = sorted(ABlist, key=lambda x: x[1])
if N % 2 == 1:
    left = Asorted[(N-1)//2][0]
    right = Bsorted[(N-1)//2][1]
    # print(left, right)
    print(right - left + 1)
else:
    if N == 2:
        left = (Asorted[(N-1)//2][0] + Asorted[(N-1)//2 + 1][0]) / 2
        right = (Bsorted[(N-1)//2][1] + Bsorted[(N-1)//2 + 1][1]) / 2
    else:
        left = (Asorted[(N-1)//2][0] + Asorted[(N-1)//2 + 2][0]) / 2
        right = (Bsorted[(N-1)//2 - 1][1] + Bsorted[(N-1)//2 + 1][1]) / 2
    print(int((right - left) * 2 + 1))