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)

def middle(a)
    (a.size % 2).zero? ? a[a.size/2 - 1, 2].inject(:+) / 2.0 : a[a.size/2]
end

def middleo(a)
    a.sort[a.size/2]
end

def middlee(a)
    a.sort[a.size/2 - 1, 2]
end

n = gets.to_i
a = (1..n).map{ gets.split(' ').map(&:to_i) }
at = a.transpose
if n.odd? then
    a1 = middleo(at[0])
    a2 = middleo(at[1])
    p ((a2 - a1).abs + 1).to_i
else
    # a1 = middle(at[0])
    # a2 = middle(at[1])
    x1 = middlee(at[0])
    x2 = middlee(at[1])
    p x1.product(x2).map{|v1,v2| (v1+v2)/ 2.0}.uniq.size
    # p ((a2 - a1)*2 + 1).to_i
end