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 OCaml (4.10.0) (judge verdict: Accepted)

let comp x y = 
    if x > y then 1 else if x < y then -1 else 0

let () = Scanf.scanf "%d\n" @@ fun n ->
    let as_ = Array.make n 0 in
    let bs_ = Array.make n 0 in
    let substitute i = 
        Scanf.scanf "%d %d\n" @@ fun a b ->
            Array.set as_ i a;
            Array.set bs_ i b;
    in
    ignore @@ Array.map substitute @@ (Array.init n @@ fun i -> i);
    ignore @@ Array.sort comp as_;
    ignore @@ Array.sort comp bs_;
    let res = 
        if n mod 2 == 0
        then
            (Array.get bs_ @@ n / 2) + (Array.get bs_ @@ n / 2 - 1)
            - (Array.get as_ @@ n / 2) - (Array.get as_ @@ n / 2 - 1) + 1
        else
            (Array.get bs_ @@ n / 2) - (Array.get as_ @@ n / 2) + 1
    in
    ignore @@ Printf.printf "%d\n" @@ res