Problem:
Score : 500 points

Problem Statement

You are given a tree with N vertices 1,2,\ldots,N, and positive integers c_1,c_2,\ldots,c_N.
The i-th edge in the tree (1 \leq i \leq N-1) connects Vertex a_i and Vertex b_i.

We will write a positive integer on each vertex in T and calculate our score as follows:

On each edge, write the smaller of the integers written on the two endpoints.

Let our score be the sum of the integers written on all the edges.

Find the maximum possible score when we write each of c_1,c_2,\ldots,c_N on one vertex in T, and show one way to achieve it. If an integer occurs multiple times in c_1,c_2,\ldots,c_N, we must use it that number of times.

Constraints

1 \leq N \leq 10000

1 \leq a_i,b_i \leq N

1 \leq c_i \leq 10^5

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
:
a_{N-1} b_{N-1}
c_1 \ldots c_N

Output

Use the following format:

M
d_1 \ldots d_N

where M is the maximum possible score, and d_i is the integer to write on Vertex i.
d_1,d_2,\ldots,d_N must be a permutation of c_1,c_2,\ldots,c_N.
If there are multiple ways to achieve the maximum score, any of them will be accepted.

Sample Input 1

5
1 2
2 3
3 4
4 5
1 2 3 4 5

Sample Output 1

10
1 2 3 4 5

If we write 1,2,3,4,5 on Vertex 1,2,3,4,5, respectively, the integers written on the four edges will be 1,2,3,4, for the score of 10. This is the maximum possible score.

Sample Input 2

5
1 2
1 3
1 4
1 5
3141 59 26 53 59

Sample Output 2

197
59 26 3141 59 53

c_1,c_2,\ldots,c_N may not be pairwise distinct.

Solution in Fortran (GNU Fortran 9.2.1) (judge verdict: Time Limit Exceeded)

program prob5
    implicit none
    integer :: N
    logical, allocatable :: G(:, :)
    integer :: a, b, i, j
    integer, allocatable :: queue(:), ans(:), c(:)
    integer :: ind, v, last, su
    logical, allocatable :: used(:)
    read(*,*) N

    allocate(G(N, N), c(N), ans(N), used(N))
    do i = 1, N
        do j = 1, N
            G(i, j) = .false.
        end do
    end do
    do i = 1, N-1
        read(*,*) a, b
        G(a, b) = .true.
        G(b, a) = .true.
    end do
    read(*,*) c
    call msort(c)

    allocate(queue(N))
    do i = 1, N
        queue(i) = 0
    end do
    ind = 1
    last = 2
    queue(1) = 1
    ans(1) = c(N)
    do i = 1, N
        used(i) = .false.
    end do
    used(1) = .true.

    do while(last <= N)
        v = queue(ind)
        do i = 1, N
            if(G(v, i) .and. .not. used(i)) then
                queue(last) = i
                last = last + 1
                used(i) = .true.
            end if
        end do
        ind = ind + 1
    end do

    do i = 1, N
        ans(queue(i)) = c(N - i + 1)
    end do
    su = 0
    do i = 1, N-1
        su = su + c(i)
    end do

    write(*,*) su
    write(*,*) ans




    
    deallocate(G, queue, c, ans, used)
    stop
contains

  !
  ! swap two arguments
  !
subroutine swap(a, b)
    implicit none
    integer, intent(inout) :: a, b

    integer :: c

    c = a
    a = b
    b = c

  end subroutine swap

  !
  ! merge sort
  !
  recursive subroutine msort(x)
    implicit none
    integer, intent(inout) :: x(:)

    integer :: n, mid, i, j, k
    integer, allocatable :: tmp(:)
    
    n = size(x)
    
    if(n == 1) then
      return
    end if

    !
    ! 前半と後半に分けてソート
    ! 1~midとmid+1~nをそれぞれソート済みにする
    !
    mid = n / 2
    call msort(x(:mid))
    call msort(x(mid+1:))

    !
    ! tmpという配列にxを一時的に保管
    ! マージソートは外部ソート
    !
    allocate(tmp(n))
    tmp = x

    ! tmpの後半部分を左右反転
    do k = n, mid+1, -1
      tmp(k) = x(mid + n - k + 1)
    end do

    ! iは前半のリストのうちまだ見ていないものの最小値
    ! jは後半のリストのうちまだ見ていないものの最小値
    ! kは見た回数の合計
    i = 1
    j = n

    do k = 1, n
      !
      ! 一番左と一番右のうち小さい方をxに入れていく
      ! ここが等号つき不等号なので、安定ソートとなる
      !
      if(tmp(i) <= tmp(j)) then
        x(k) = tmp(i)
        i = i + 1
      else
        x(k) = tmp(j)
        j = j - 1
      end if
    end do
    
    deallocate(tmp)
    return
  end subroutine msort

end program prob5