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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;
typedef int64_t i64;

static void print_int(i64 n){if(n<0){putchar('-');n=-n;}if(n==0){putchar('0');return;}int s[20],len=0;while(n>0){s[len++]=n%10+'0';n/=10;}while(len>0){putchar(s[--len]);}}
static i64 read_int(void){int prev='\0';int c=getchar();while(!('0'<=c && c<='9')){prev=c;c=getchar();}i64 res=0;while('0'<=c && c<='9'){res=10*res+c-'0';c=getchar();}return prev=='-'?-res:res;}

typedef struct directed_edge {
  int32_t vertex;
  int32_t next;
} graph_edge;

typedef struct directedGraph {
  graph_edge *edge;
  int32_t *start;
  int32_t pointer;
  int32_t vertex_num;
  int32_t edge_max_size;
} graph;

graph* new_graph (const int vertex_num) {
  graph *g = (graph *) calloc (1, sizeof (graph));
  g->edge = (graph_edge *) calloc (1, sizeof (graph_edge));
  g->start = (int32_t *) calloc (vertex_num, sizeof (int32_t));
  g->pointer = 0;
  g->vertex_num = vertex_num;
  g->edge_max_size = 1;
  for (int32_t i = 0; i < vertex_num; ++i) {
    g->start[i] = -1;
  }
  return g;
}

void add_edge (graph *g, int32_t from, int32_t to) {
  if (g->pointer == g->edge_max_size) {
    g->edge_max_size *= 2;
    g->edge = (graph_edge *) realloc (g->edge, sizeof (graph_edge) * g->edge_max_size);
  }
  g->edge[g->pointer] = (graph_edge) {to, g->start[from]};
  g->start[from] = g->pointer++;
}

void BFS_graph (graph *g, int32_t src, int32_t *q) {
  uint8_t *used = (uint8_t *) calloc (g->vertex_num, sizeof (uint8_t));
  int32_t front = 0;
  int32_t last = 0;
  used[src] = 1;
  q[last++] = src;
  while (front < last) {
    const int32_t v = q[front++];
    for (int32_t p = g->start[v]; p != -1; p = g->edge[p].next) {
      const int32_t u = g->edge[p].vertex;
      if (!used[u]) {
	used[u] = 1;
	q[last++] = u;
      }
    }
  }
  free(used);
}

void run (void) {
  i32 n = read_int();
  graph *g = new_graph (n);
  for (i32 i = 1; i < n; ++i) {
    i32 a = read_int() - 1;
    i32 b = read_int() - 1;
    add_edge (g, a, b);
    add_edge (g, b, a);
  }
  const i32 m = 100000;
  i32 *cnt = (i32 *) calloc (m + 1, sizeof (i32));
  for (i32 i = 0; i < n; ++i) {
    cnt[read_int()]++;
  }
  i32 *q = (i32 *) calloc (n, sizeof (i32));
  BFS_graph (g, 0, q);
  i32 *ans = (i32 *) calloc (n, sizeof (i32));
  i32 sum = 0;
  i32 k = 1;
  for (i32 i = n - 1; i >= 0; --i) {
    for (; cnt[k] == 0; ++k);
    i32 v = q[i];
    ans[v] = k;
    cnt[k]--;
    if (i > 0) sum += k;
  }
  print_int (sum);
  putchar ('\n');
  for (i32 i = 0; i < n; ++i) {
    print_int (ans[i]);
    putchar (i == n - 1 ? '\n' : ' ');
  }
}

int main (void) {
  run();
  return 0;
}