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 Ruby (2.7.1) (judge verdict: Accepted)

n = gets.chomp.to_i

a_arr = []
b_arr = []
n.times do
  a, b = gets.chomp.split(" ")
  a_arr << a.to_i
  b_arr << b.to_i
end

a_arr.sort!
b_arr.sort!

if n % 2 == 0
  median_a = (a_arr[n / 2 - 1] + a_arr[n / 2]).to_f / 2
  median_b = (b_arr[n / 2 - 1] + b_arr[n / 2]).to_f / 2
  puts ((median_b - median_a) * 2 + 1).to_i
else
  median_a = a_arr[(n - 1)/ 2]
  median_b = b_arr[(n - 1)/ 2]
  puts median_b - median_a + 1
end