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 PyPy3 (2.4.0) (judge verdict: Accepted)

from collections import Counter,defaultdict,deque
import sys
import bisect
import math
import itertools
import string
import queue
import copy
# import numpy as np
# import scipy
from itertools import permutations, combinations
from heapq import heappop, heappush
input = sys.stdin.readline
sys.setrecursionlimit(10**8)
mod = 10**9+7

def inp(): # n=1
    return int(input())
def inpm(): # x=1,y=2
    return map(int,input().split())
def inpl(): # a=[1,2,3,4,5,...,n]
    return list(map(int, input().split()))
def inpls(): # a=['1','2','3',...,'n']
    return list(input().split())
def inplm(n): # x=[] 複数行
    return list(int(input()) for _ in range(n))
def inpll(n): # [[1,1,1,1],[2,2,2,2],[3,3,3,3]]
    return sorted([list(map(int, input().split())) for _ in range(n)])
def sortx(x,n,k):
    if k == 0:x.sort(key=lambda y:y[1,n])
    else:x.sort(reversed=True, key=lambda y:y[1,n])

def graphm():
    n,m=inpm()
    g=[[] for _ in range(n)]
    for _ in range(m):
        a,b,w=inpm()
        a-=1
        b-=1
        g[a].append((w,b))
        g[b].append((w,a))
    return n,m,g

def warshall_floyd(): # d[i][j]: iからjへの最短距離
    n,m,r = map(int,input().split()) #n:頂点数　w:辺の数
    R=inpl()
    d=[[10**10 for _ in range(n)] for _ in range(n)]
    for i in range(m):
        x,y,z = inpm()
        x -= 1
        y -= 1
        d[x][y] = z
        d[y][x] = z
    for i in range(n):
        d[i][i] = 0 #自身のところに行くコストは0
    for k in range(n):
        for i in range(n):
            for j in range(n):
                d[i][j] = min(d[i][j],d[i][k] + d[k][j])
    ans = float('inf')
    for com in permutations(R,r):
        ans_pre = 0
        for i in range(r-1):
            ans_pre += d[com[i]-1][com[i+1]-1]
        ans = min(ans,ans_pre)
    print(ans)

def main(): 
    warshall_floyd()

if __name__ == "__main__":
    main()