Problem:
Score : 600 points

Problem Statement

There is a function f(x), which is initially a constant function f(x) = 0.

We will ask you to process Q queries in order. There are two kinds of queries, update queries and evaluation queries, as follows:

An update query 1 a b: Given two integers a and b, let g(x) = f(x) + |x - a| + b and replace f(x) with g(x).

An evaluation query 2: Print x that minimizes f(x), and the minimum value of f(x). If there are multiple such values of x, choose the minimum such value.

We can show that the values to be output in an evaluation query are always integers, so we ask you to print those values as integers without decimal points.

Constraints

All values in input are integers.

1 \leq Q \leq 2 \times 10^5

-10^9 \leq a, b \leq 10^9

The first query is an update query.

Input

Input is given from Standard Input in the following format:

Q
Query_1
:
Query_Q

See Sample Input 1 for an example.

Output

For each evaluation query, print a line containing the response, in the order in which the queries are given.

The response to each evaluation query should be the minimum value of x that minimizes f(x), and the minimum value of f(x), in this order, with space in between.

Sample Input 1

4
1 4 2
2
1 1 -8
2

Sample Output 1

4 2
1 -3

In the first evaluation query, f(x) = |x - 4| + 2, which attains the minimum value of 2 at x = 4.

In the second evaluation query, f(x) = |x - 1| + |x - 4| - 6, which attains the minimum value of -3 when 1 \leq x \leq 4. Among the multiple values of x that minimize f(x), we ask you to print the minimum, that is, 1.

Sample Input 2

4
1 -1000000000 1000000000
1 -1000000000 1000000000
1 -1000000000 1000000000
2

Sample Output 2

-1000000000 3000000000

Solution in Fortran (gfortran v4.8.4) (judge verdict: Accepted)

module mod_priority_queue
  implicit none
  type t_priority_queue
    private
    integer :: num = 0 !中に入ってる物の数
    integer, pointer :: heap(:) => null()!pqの容量
  contains
    procedure :: offer => offer
    procedure :: clear => clear
    procedure :: poll => poll
    procedure :: peek => peek
    procedure :: size => size_of
  end type t_priority_queue
contains
  subroutine offer(pq,item)
    class(t_priority_queue), intent(inout) :: pq
    integer, intent(in) :: item
    integer :: n, i, t
    integer, allocatable :: tmp(:)
    if (.not.associated(pq%heap)) allocate(pq%heap(1))

    !いっぱいいっぱいのときには拡張している
    if (pq%num == size(pq%heap)) then
      allocate(tmp(pq%num))
      tmp = pq%heap
      deallocate(pq%heap)
      allocate(pq%heap(2*pq%num))
      pq%heap(1:pq%num) = tmp
      deallocate(tmp)
    end if

    pq%num = pq%num+1
    pq%heap(pq%num) = item
    n = pq%num
    do while (n > 1)
      i = n/2
      if (pq%heap(n) > pq%heap(i)) then
        t = pq%heap(n)
        pq%heap(n) = pq%heap(i)
        pq%heap(i) = t
      end if
      n = i
    end do
    return
  end subroutine offer

  subroutine clear(pq)
    class(t_priority_queue), intent(inout) :: pq
    if (associated(pq%heap)) deallocate(pq%heap)
    pq%num = 0
    return
  end subroutine clear

  function poll(pq) result(item)
    class(t_priority_queue), intent(inout) :: pq
    integer :: item, n, i, j, tmp
    n = pq%num
    item = pq%heap(1)
    pq%heap(1) = pq%heap(n)
    pq%num = pq%num-1
    i = 1
    do while (2*i < n)
      j = 2*i
      if (j+1 < n .and. pq%heap(j+1) > pq%heap(j)) j = j+1
      if (pq%heap(j) > pq%heap(i)) then
        tmp = pq%heap(j)
        pq%heap(j) = pq%heap(i)
        pq%heap(i) = tmp
      end if
      i = j
    end do
  end function poll
  subroutine pop(pq)
    class(t_priority_queue), intent(inout) :: pq
    integer :: item, n, i, j, tmp
    n = pq%num
    item = pq%heap(1)
    pq%heap(1) = pq%heap(n)
    pq%num = pq%num-1
    i = 1
    do while (2*i < n)
      j = 2*i
      if (j+1 < n .and. pq%heap(j+1) > pq%heap(j)) j = j+1
      if (pq%heap(j) > pq%heap(i)) then
        tmp = pq%heap(j)
        pq%heap(j) = pq%heap(i)
        pq%heap(i) = tmp
      end if
      i = j
    end do
  end subroutine 
    
  function peek(pq) result(item)
    class(t_priority_queue), intent(inout) :: pq
    integer :: item
    item = pq%heap(1)
  end function peek

  integer function size_of(pq)
    class(t_priority_queue), intent(in) :: pq
    size_of = pq%num
  end
end module mod_priority_queue

program am
use mod_priority_queue
implicit none
integer::Q
integer::T,A(2)
integer(16)::ANS

type(t_priority_queue)::L,R
integer::i

read*,Q
ANS=0
call offer(L,-huge(i))
call offer(R,-huge(i))
do i=1,Q
    read*,T,A(2*T-1:2)
    select case(T)
    case(1)
        ANS=ANS+A(2)
        if(A(1)<peek(L))then
            ANS=ANS+peek(L)-A(1)
        else if(-peek(R)<A(1))then
            ANS=ANS+A(1)+peek(R)
        endif
        call offer(l,A(1))
        call offer(r,-A(1))
        if(peek(L)>-peek(R))then
            call offer(l,-peek(R))
            call offer(r,-peek(l))
            call pop(l)
            call pop(r)
        endif
    case(2)
        print"(i0,a,i0)",peek(l)," ",ANS
    end select
end do
end program am