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 Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

program e
    use,intrinsic :: iso_fortran_env
    implicit none
    integer(int32):: n,i
    integer(int32), allocatable:: a(:),b(:)

    read*, n
    allocate(a(n),b(n))

    do i=1,n
        read*, a(i),b(i)
    end do
    call merge_sort(a,1,n)
    call merge_sort(b,1,n)

    if (mod(n,2)==1) then
        print'(i0)', b(n/2+1) - a(n/2+1) + 1
    else
        print'(i0)', b(n)+b(n-1) - (a(1)+a(2))+1
    end if
        
contains
        recursive subroutine merge_sort(ar, fst, lst)
        integer(int32),intent(inout):: ar(:)
        integer(int32),intent(in):: fst,lst
        integer(int32):: mdl

        if (lst-fst < 2) then
            if (ar(fst) > ar(lst)) call swap(ar(fst),ar(lst))
            return
        end if

        mdl = (fst+lst)/2
        call merge_sort(ar, fst, mdl)
        call merge_sort(ar, mdl+1, lst)
        call merge_(ar, fst, mdl, lst)
    end subroutine

    subroutine merge_(ar, fst, mdl, lst)
        integer(int32),intent(inout):: ar(:)
        integer(int32),intent(in):: fst, mdl, lst
        integer(int32),allocatable:: tmp(:)
        integer(int32):: li, ri, ti

        allocate(tmp(lst-fst+1))

        li=fst
        ri=mdl+1   
        ti=1

        do while (li <= mdl .and. ri <= lst)
            if (ar(li) <= ar(ri)) then
                tmp(ti) = ar(li)
                li=li+1
            else
                tmp(ti) = ar(ri)
                ri=ri+1
            end if
            ti=ti+1
        end do

        if (li <= mdl) then
            tmp(ti:) = ar(li:mdl)
        else
            tmp(ti:) = ar(ri:lst)
        end if

        ar(fst:lst) = tmp(:)
        deallocate(tmp)
    end subroutine

    subroutine swap(x,y)
        integer(int32),intent(inout):: x,y
        integer(int32):: tmp
        tmp = x
        x = y
        y = tmp
    end subroutine
end program e