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: Wrong Answer)

n = gets.to_i
num = []
n.times{ num << gets.split&.map(&:to_i)}

l_num = []
r_num = []


num.map do |n|
  l_num << n[0]
  r_num << n[1]
end

def middle(array)
  sa = array.sort
  if sa.size.even?
   (sa[sa.size/2] + sa[sa.size/2 - 1] )/ 2.0
  else
    sa[sa.size/2]
  end
end

l_num_middle = middle(l_num)
r_num_middle = middle(r_num)

if n.even?
  p (r_num_middle - l_num_middle) * 2 + 1
else
  p r_num_middle - l_num_middle + 1
end