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 Lua (LuaJIT 2.1.0) (judge verdict: Time Limit Exceeded)

local mfl, mce = math.floor, math.ceil
local n, m, r = io.read("*n", "*n", "*n")
local t = {}
for i = 1, r do
  t[i] = io.read("*n")
end
local edge = {}
for i = 1, n do
  edge[i] = {}
end
for i = 1, m do
  local a, b, c = io.read("*n", "*n", "*n")
  edge[a][b], edge[b][a] = c, c
end
local inf = 1000000007

local function getlength(start_idx, dst_idx)
  local taskstate = {}
  for i = 1, n do taskstate[i] = false end
  local tasks = {}
  local tasknum = 0
  local done = 0

  local len = {}
  for i = 1, n do len[i] = inf end
  len[start_idx] = 0

  local function addtask(idx)
    if not taskstate[idx] then
      taskstate[idx] = true
      tasknum = tasknum + 1
      tasks[tasknum] = idx
    end
  end
  addtask(start_idx)

  while done < tasknum do
    done = done + 1
    local src = tasks[done]
    taskstate[src] = false
    for dst, cost in pairs(edge[src]) do
      if len[src] + cost < len[dst] then
        len[dst] = len[src] + cost
        addtask(dst)
      end
    end
  end
  return len[dst_idx]
end

local ret = 1000000007 * 1000

local function getpattern(n, patall, idx)
  local used = {}
  local retary = {}
  local div = patall
  for i = 1, n do used[i] = false end
  for i = n, 1, -1 do
    div = mfl(div / i)
    local v_idx = mfl(idx / div)
    idx = idx % div
    local tmp_idx = 0
    for j = 1, n do
      if not used[j] then
        if tmp_idx == v_idx then
          table.insert(retary, j)
          used[j] = true
          break
        else
          tmp_idx = tmp_idx + 1
        end
      end
    end
  end
  return retary
end
local tot = 1
for i = 1, r do
  tot = tot * i
end
for i = 0, tot - 1 do
  local pat = getpattern(r, tot, i)
  local z = 0
  for i = 1, r - 1 do
    z = z + getlength(t[pat[i]], t[pat[i + 1]])
  end
  if z < ret then ret = z end
end
print(ret)