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)

require 'set'; require 'prime'; require 'pp'
require 'bigdecimal'
INF=Float::INFINITY; MOD=10**9+7
n = gets.chomp.to_i
a, b = n.times.map{gets.split.map(&:to_i)}.transpose

if n.odd?
  # p [a[n/2], b[n/2]]
  puts b[n/2] - a[n/2] + 1
else

  amid = BigDecimal((a[n/2-1] + a[n/2]).to_s) / 2
  bmid = BigDecimal((b[n/2-1] + b[n/2]).to_s) / 2
  # p [amid, bmid]
  ret = (bmid - amid + BigDecimal("0.5")) * 2
  puts ret.to_i
end