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 Rust (1.42.0) (judge verdict: Wrong Answer)

macro_rules !read_value {($iter :expr ,($($t :tt ) ,*) ) =>{($(read_value !($iter ,$t ) ) ,*) } ;($iter :expr ,[$t :tt ;$len :expr ] ) =>{(0 ..$len ) .map (|_ |read_value !($iter ,$t ) ) .collect ::<Vec <_ >>() } ;($iter :expr ,{chars :$base :expr } ) =>{read_value !($iter ,String ) .chars () .map (|c |(c as u8 -$base as u8 ) as usize ) .collect ::<Vec <usize >>() } ;($iter :expr ,{char :$base :expr } ) =>{read_value !($iter ,{chars :$base } ) [0 ] } ;($iter :expr ,chars ) =>{read_value !($iter ,String ) .chars () .collect ::<Vec <char >>() } ;($iter :expr ,char ) =>{read_value !($iter ,chars ) [0 ] } ;($iter :expr ,usize1 ) =>{read_value !($iter ,usize ) -1 } ;($iter :expr ,$t :ty ) =>{$iter .next () .unwrap () .parse ::<$t >() .unwrap () } ;}
macro_rules !input_inner {($iter :expr ) =>{} ;($iter :expr ,) =>{} ;($iter :expr ,mut $var :ident :$t :tt $($r :tt ) *) =>{let mut $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,mut $var :ident $($r :tt ) *) =>{input_inner !{$iter ,mut $var :usize $($r ) *} } ;($iter :expr ,$var :ident :$t :tt $($r :tt ) *) =>{let $var =read_value !($iter ,$t ) ;input_inner !{$iter $($r ) *} } ;($iter :expr ,$var :ident $($r :tt ) *) =>{input_inner !{$iter ,$var :usize $($r ) *} } ;}
macro_rules !input {(source =$s :expr ,$($r :tt ) *) =>{let mut iter =$s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;(iter =$iter :ident ,$($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut $iter =s .split_whitespace () ;input_inner !{$iter ,$($r ) *} } ;($($r :tt ) *) =>{let s ={use std ::io ::Read ;let mut s =String ::new () ;std ::io ::stdin () .read_to_string (&mut s ) .unwrap () ;s } ;let mut iter =s .split_whitespace () ;input_inner !{iter ,$($r ) *} } ;}
fn main() {
    input!(n, ab: [(usize, usize); n]);
    let (mut a, mut b): (Vec<_>, Vec<_>) = ab.iter().cloned().unzip();
    a.sort();
    b.sort();
    if n % 2 == 1 {
        let l = a[n / 2];
        let u = b[n / 2];
        println!("{}", u - l + 1);
    } else {
        let l2 = a[n / 2];
        let l1 = a[n / 2 - 1];
        let u2 = b[n / 2];
        let u1 = b[n / 2 - 1];
        let l = if l1 == l2 { l1 + l2 } else { l2 * 2 - 1 };
        let u = if u1 == u2 { u1 + u2 } else { u1 * 2 + 1 };
        println!("{}", u - l + 1);
    }
}