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)

n = int(input())
A = [-1] * n
B = [-1] * n
for i in range(n):
    A[i], B[i] = map(int, input().split())

A.sort()
B.sort()

if n % 2 == 1:
    avrA = A[n // 2]
    avrB = B[n // 2]

    print(avrB - avrA + 1)
else:
    avr_low_A = A[n // 2 - 1]
    avr_high_A = A[n // 2]
    avr_low_B = B[n // 2 - 1]
    avr_high_B = B[n // 2]

#     x = avr_low_B - avr_low_A + 1
#     y = avr_high_B - avr_high_A + 1
#     sum = 0
#     if x > y:
#         x, y = y, x
#     for i in range(2, x + 1):
#         m = (y - x + 1) * x
#         sum += x + y - m + 1
#
    print((avr_high_A + avr_high_B) - (avr_low_A + avr_low_B) + 1)