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.to_i
ABS = N.times.map { gets.split.map(&:to_i) }

as = []
bs = []
ABS.each do |ab|
  a, b = ab
  as.push(a)
  bs.push(b)
end

def m(arr)
  sorted = arr.sort
  if arr.count % 2 == 1
    sorted[(arr.count + 1) / 2 - 1]
  else
    (sorted[arr.count / 2 - 1] + sorted[arr.count / 2 + 1 - 1]) / 2.0
  end
end

am = m(as)
bm = m(bs)

if N % 2 == 1
  puts (bm - am + 1).round
else
  puts ((bm - am) * 2 + 1).round
end