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 (2.7.1) (judge verdict: Accepted)

class PriorityQueue
  class Node
    attr_reader :value, :priority
    def initialize(value, priority)
      @value = value
      @priority = priority
    end
  end

  def initialize
    @heap = [nil]
  end

  def size
    @heap.size - 1
  end

  def empty?
    size == 0
  end

  def enqueue(value, priority)
    @heap.push(Node.new(value, priority))
    up_heap
    self
  end

  def dequeue
    return nil if empty?
    return @heap.pop.value if size == 1
    ret = @heap[1].value
    @heap[1] = @heap.pop
    down_heap
    ret
  end

  def peek
    @heap[1].value.dup
  end

  private

  def up_heap
    pos = size
    while pos > 1
      parent = pos / 2
      break if @heap[pos].priority <= @heap[parent].priority
      tmp = @heap[pos]
      @heap[pos] = @heap[parent]
      @heap[parent] = tmp
      pos = parent
    end
  end

  def down_heap
    pos = 1
    loop{
      left = pos * 2 
      right = pos * 2 + 1
      break if left >= @heap.size
      if right >= @heap.size || @heap[left].priority >= @heap[right].priority
        break if @heap[pos].priority >= @heap[left].priority
        tmp = @heap[pos]
        @heap[pos] = @heap[left]
        @heap[left] = tmp
        pos = left
      else
        break if @heap[pos].priority >= @heap[right].priority
        tmp = @heap[pos]
        @heap[pos] = @heap[right]
        @heap[right] = tmp
        pos = right
      end
    }
  end
end

N,M,R=gets.split.map(&:to_i)
r=gets.split.map(&:to_i)
E=Array.new(N+1){[]}
M.times{
	a,b,c=gets.split.map(&:to_i)
	E[a]<<[b,c]
	E[b]<<[a,c]
}
warn E.inspect
D=[]
r.each{|s|
	d=Array.new(N+1){10**9}
	d[s]=0
	q=PriorityQueue.new
	q.enqueue(s,-d[s])
	until q.empty?
		v=q.dequeue
		E[v].each{|u,l|
			if d[u]>d[v]+l
				d[u]=d[v]+l
				q.enqueue(u,-d[u])
			end
		}
	end
	D[s]=d
}
warn D.inspect
q=PriorityQueue.new
r.each{|s|
	q.enqueue([[s],0],0)
}
min = 10**9
until q.empty?
	v,d=q.dequeue
	if v.size == r.size
		min=d
		warn v.inspect
		break
	end
	r.each{|o|
		next if v.include?(o)
		n=v+[o]
		nd=d+D[v[-1]][o]
		q.enqueue([n,nd],-nd)
	}
end
puts min