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 Scala (2.11.7) (judge verdict: Accepted)

object Main {

  def main(args: Array[String]): Unit = {
    import scala.collection.immutable.Queue
    import scala.collection.mutable.ArrayBuffer
    val sc = new java.util.Scanner(System.in)

    val N = sc.nextShort()
    val E = Array.fill(N)(ArrayBuffer[Short]())
    val Dim = new Array[Int](N)
    val visited = new Array[Boolean](N)
    (0 until N - 1) foreach { _ =>
      val _a, _b = sc.nextShort()
      val a: Short = (_a - 1).toShort
      val b: Short = (_b - 1).toShort
      E(a).append(b)
      E(b).append(a)
      Dim(a) += 1
      Dim(b) += 1
    }
    var C = Array.fill(N)(sc.nextInt()).sorted(Ordering.Int.reverse).toList
    val D = new Array[Int](N) // 最終結果
    // 入次数が小さい順に訪れて行き,

    var M = 0L
    val (_, root) = Dim.zipWithIndex.minBy(_._1) // 入次数がもっとも小さいものをルートとする
    def bfs(queue: Queue[Short]): Unit =
      if (queue.nonEmpty) {
        val (u, dequeued) = queue.dequeue
        val finded = E(u) collect { case v if !visited(v) =>
          visited(v) = true
          D(v) = C.head // 一番価値が高いものを取り出してuに入れる
          C = C.tail // 残りをCに入れて更新する
          M += D(v) // 新しく追加した辺の価値は子のほうがちいさいので蓄積
          v // さらにその子孫を探索
        }
        bfs(finded.foldLeft(dequeued)((b,n) => b.enqueue(n)))
      } else {}

    visited(root) = true
    D(root) = C.head
    C = C.tail
    bfs(Queue[Short](root.toShort))

    println(M)
    println(D.mkString(" "))
  }

}