Problem:
Score : 400 points

Problem Statement

There are N towns in the State of Atcoder, connected by M bidirectional roads.

The i-th road connects Town A_i and B_i and has a length of C_i.

Joisino is visiting R towns in the state, r_1,r_2,..,r_R (not necessarily in this order).

She will fly to the first town she visits, and fly back from the last town she visits, but for the rest of the trip she will have to travel by road.

If she visits the towns in the order that minimizes the distance traveled by road, what will that distance be?

Constraints

2≤N≤200

1≤M≤N×(N-1)/2

2≤R≤min(8,N) (min(8,N) is the smaller of 8 and N.)

r_i≠r_j (i≠j)

1≤A_i,B_i≤N, A_i≠B_i

(A_i,B_i)≠(A_j,B_j),(A_i,B_i)≠(B_j,A_j) (i≠j)

1≤C_i≤100000

Every town can be reached from every town by road.

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M R
r_1 ... r_R
A_1 B_1 C_1
:
A_M B_M C_M

Output

Print the distance traveled by road if Joisino visits the towns in the order that minimizes it.

Sample Input 1

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

Sample Output 1

2

For example, if she visits the towns in the order of 1, 2, 3, the distance traveled will be 2, which is the minimum possible.

Sample Input 2

3 3 2
1 3
2 3 2
1 3 6
1 2 2

Sample Output 2

4

The shortest distance between Towns 1 and 3 is 4. Thus, whether she visits Town 1 or 3 first, the distance traveled will be 4.

Sample Input 3

4 6 3
2 3 4
1 2 4
2 3 3
4 3 1
1 4 1
4 2 2
3 1 6

Sample Output 3

3

Solution in Ruby (2.3.3) (judge verdict: Wrong Answer)

# coding: utf-8
#
# ----------------------
# WarshallFloyd
# ----------------------
# :u ノードのインデックス
# :k u の出次数
# :v u に隣接する頂点の番号
# :w 隣接する頂点への重み
#
INF  = Float::INFINITY
Node = Struct.new(:u, :k, :v, :w)

class Graph
  attr :nodes, :dist
  def initialize(n)
    @nodes = Array.new(n){ Node.new }
    @nodes = @nodes.map.with_index do |node,idx|
      node.u = idx
      node
    end
    @dist = Array.new(n).map{Array.new(n, 0)}
  end

  def add_graph_edge(u, v, w)
    #puts "#{u} -> #{v} (#{w})"
    if @nodes[u].v.nil?
      @nodes[u].v = [v]
      @nodes[u].w = {v=>w}
    elsif @nodes[u].v.is_a?(Array)
      @nodes[u].v << v
      @nodes[u].w = {v=>w}.merge(@nodes[u].w)
    else
      @nodes[u].v = [@nodes[u].v, v]
      @nodes[u].w = {v=>w}.merge(@nodes[u].w)
    end
    @nodes[u].k = if @nodes[u].v.is_a?(Array)
                    @nodes[u].v.length
                  else
                    1
                  end
  end

  def warshall_floyd()
    # initialize
    for i in 0...@nodes.length
      for j in 0...@nodes.length
        @dist[i][j] = if not @nodes[i].w.nil? and @nodes[i].w.has_key?(j)
                        @nodes[i].w[j]
                      else
                        INF
                      end
      end
      @dist[i][i] = 0
    end

    # calc
    for k in 0...@nodes.length
      for i in 0...@nodes.length
        for j in 0...@nodes.length
          if @dist[i][k] != INF and @dist[k][j] != INF
            @dist[i][j] = [@dist[i][j], @dist[i][k] + @dist[k][j]].min
          end
        end
      end
    end
  end

  def has_negative_cycle?()
    ans = false
    for v in 0...V
      ans = true if @dist[v][v] < 0
    end
    ans
  end
end

lines = $stdin.read
array = lines.split("\n")

N,M,R = array[0].split(" ").map(&:to_i)
$graph = Graph.new(N)
$rarr  = array[1].split(" ").map(&:to_i).map{|idx| idx-1}

array[2..M+1].each do |s|
  s,t,d = s.split(" ").map(&:to_i)
  s,t=s-1,t-1
  $graph.add_graph_edge(s,t,d)
  $graph.add_graph_edge(t,s,d)
end

# execute WarshallFloyd !
$graph.warshall_floyd()
ans = INF

#$rarr.permutation(R).each do |picked|
#  s = 0
#  p picked
#  picked.each_cons(2).each do |arr|
#    s += $graph.dist[arr.first][arr.last]
#  end
#  ans = [s,ans].min
#end

puts ans