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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Linq;
using System.Collections.Generic;
using static System.Console;
using System.Text;

namespace M_SOLUTIONS
{
    class Program
    {
        static public int[] Sarray() { return ReadLine().Split().Select(int.Parse).ToArray(); }
        static public List<int> Slist() { return ReadLine().Split().Select(int.Parse).ToList(); }
        static void Main(string[] args)
        {
            var n = Sarray()[0];
            var grf = new Graph(n);
        }
        public class Graph
        {
            List<int>[] graph;
            int[] v;
            class edgenum
            {
                public int V { get; }//頂点
                public int N { get; set; }//枝の数
                public edgenum(int v, int n) { V = v;N = n; }
            }
            public Graph(int n)//無向グラフ
            {
                graph = new List<int>[n];
                for (int i = 0; i < n; ++i)
                    graph[i] = new List<int>();
                v = new int[n];

                //入力方式に従って適当に
                for (int i = 0; i < n-1; ++i)
                {
                    var info = Sarray();//ai, bi
                    var ai = info[0]-1;
                    var bi = info[1]-1;
                    graph[ai].Add(bi);
                    graph[bi].Add(ai);
                }
                val = new int[n];
                var ci = Slist();
                ci.Sort((l, r) => r - l);
                //頂点の枝の数
                var ve = new List<edgenum>();
                var vefuhen = new List<edgenum>();
                for (var i = 0; i < n; ++i)
                {
                    var egn = new edgenum(i, graph[i].Count);
                    ve.Add(egn);
                    vefuhen.Add(egn);
                }
                ve.Sort((l, r) => r.N - l.N);
                var root = ve[0].V;

                int nIdx = 0;
                while (nIdx < n)
                {
                    ve[0].N = 0;
                    if (0 == val[ve[0].V])
                    { val[ve[0].V] = ci[nIdx++]; }//頂点の値を設定
                    foreach (var t in graph[ve[0].V])
                    {
                        vefuhen[t].N -= 1;
                        if (0 == val[t])
                        { val[t] = ci[nIdx++]; }
                    }
                    ve.Sort((l, r) => r.N - l.N);
                }
                dfs(root);
                WriteLine(cost);
                StringBuilder str = new StringBuilder();
                for(var i=0;i<n;++i)
                {
                    str.Append($"{val[i].ToString()} ");
                }
                WriteLine(str.ToString().Trim());
                
            }
            int cost = 0;
            int[] val;

            public void dfs(int next)
            {
                if (0 != v[next]) return;
                v[next] = 1;
                foreach (var e in graph[next])
                {
                    if (v[e] == 0)
                    { cost += Math.Min(val[next], val[e]); }
                    dfs(e);
                }
            }
        }
    }
}