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: Accepted)

def calcs(x,y,ax,ay,bx,by) ((ax-x)*(by-y)-(bx-x)*(ay-y))/2.0 end
def inpf() a=gets.chomp.split(" ").map(&:to_f)end
def inps() a=gets.chomp.split(" ")end  
def copy(a) Marshal.load(Marshal.dump(a)) end
def factorial(n)(n < 2)? 1 : (2..n).inject(:*) end
def scount(a) b=na(a.max+1);a.each{|n|b[n]+=1};return b end
def na(n,d=0) Array.new(n,d)end
def na2(n,m,d=0) Array.new(n){Array.new(m,d)}end
def na3(n,m,l,d=0) Array.new(n){Array.new(m){Array.new(l,d)}}end
def yn(f) return (f)? "Yes": "No" end
def ynb(f) return (f)? "YES": "NO" end
def ap(a) a.each{|u| p u} end
def bit(n) n.to_s(2).split("").map(&:to_i) end
def inp() a=gets.chomp.split(" ").map(&:to_i)end
  
def ser(sum,d,bef,used)
  if(d == @size - 1)
    @ans = [sum,@ans].min
    return
  end
  @size.times do |i|
    next if used[i] 
    used[i] = true
    ser(sum+@rt[bef][@r[i]],d+1,i,used)
    used[i] = false
  end
  
end
n,m,r = inp
r = inp
@ans = 1e20
@r = r
@size = r.size
d = []
path = na2(n+1,0)
m.times do|i|
  a,b,c = inp
  path[a].push([b,c])
  path[b].push([a,c])
end
@rt = []
r.each do |start|
  dist = na(n+1,1e20)
  dist[start] = 0
  dc = 0
  used = na(n+1,false)
  while dc < n
    i = nil
    min = 1e20
    (1..n).each do |j|
      next if used[j]
      if dist[j] <= min
        i = j
        min = dist[j]
      end
    end
    used[i] = true
    path[i].each do |nex,path_dist|
      next if used[nex]
      dist[nex] = [dist[nex],dist[i]+path_dist].min
    end
    dc += 1
  end
  @rt.push(copy(dist))
end
@size.times do |i|
  used = na(@size,false)
  used[i] = true
  ser(0,0,i,used)
  used[i] = false
end
p @ans
=begin

=end