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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<math.h>
#include<queue>
#include<deque>
#include<stack>
#include<cstdio>
#include<utility>
#include<set>
#include<list>
#include<cmath>
#include<stdio.h>
#include<string.h>
#include<iomanip>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
#define FOR(i, a, b) for (ll i = (a); i <= (b); i++)
#define REP(i, n) FOR(i, 0, n - 1)
#define NREP(i, n) FOR(i, 1, n)
using ll = long long;
using pii = pair<int, int>;
using piii = pair<pii, pii>;
const ll dx[4] = { -1, 0, 0, 1 };
const ll dy[4] = { 0, -1, 1, 0 };
const ll INF = 1e9 + 7;
const ll inf = 1LL << 50;
int gcd(int x, int y) {
	if (x < y)swap(x, y);
	if (y == 0)return x;
	return gcd(y, x%y);
}
void mul(ll a, ll b) {
	a = a * b % INF;
}
struct Accelerate_Cin {
	Accelerate_Cin() {
		cin.tie(0);
		ios::sync_with_stdio(0);
		cout << fixed << setprecision(20);
	};
};
///////////////////////////////////////

//あらかじめソートすることでc1>=c2>=,,,,>=cNとしてよい
//それらの辺とその端点からなるTの部分グラフをUとするUは森になるのでk+1個以上の頂点を持つ
//したがってk本の辺に書き込まれる整数のうち最小のものはck+1以下になる。
//この考察により各辺に書き込まれる整数をx1,,,xN-1するとxi<=ci+1となることがわかる
//したがってスコアは常にc2+,,,cN以下
//またスコアはc2+,,,,+cNとなるように構成できる
//Tの辺を選んで両端にc1,c2をそれぞれ書き込む
//k=2,,,,N-1に対して次の操作を繰り返す
//e1,,ekとその端点からなるTの部分グラフが連結になるようにTの辺ekを選ぶことができる

int N;
vector<vector<int>>V;
vector<int>c;
vector<pii>key;
int ans[10010];
bool visited[10010];
int cnt = 0;
void dfs(int a) {
	if (visited[a])return;
	ans[a] = c[cnt];
	cnt++;
	visited[a] = true;
	for (int x : V[a]) {
		dfs(x);
	}
}
int main() {
	cin >> N;
	V.resize(N);
	int sum = 0;
	REP(i, N - 1) {
		int a, b; cin >> a >> b;
		a--, b--;
		V[a].push_back(b);
		V[b].push_back(a);
	}
	key.resize(N);
	for (int i = 0; i < N; ++i) {
		key[i] = { V[i].size(),i };
	}
	sort(key.begin(), key.end(), greater<pii>());
	c.resize(N);
	REP(i, N) {
		cin >> c[i];
		sum += c[i];
		visited[i] = false;
	}
	sort(c.begin(), c.end(),greater<int>());
	sum -= c[0];
	cout << sum << endl;
	dfs(0);
	REP(i, N)cout << ans[i] << " ";
	return 0;
}