"""

bitdpっぽくやるのね…
10^7...
厳しめ

dp[bit][最後][LorR] = mincost

"""

from sys import stdin

import heapq
def Dijkstra(lis,start):

    ret = [float("inf")] * len(lis)
    ret[start] = 0
    q = [(0,start)]

    while len(q) > 0:
        ncost,now = heapq.heappop(q)
        if ncost != ret[now]:
            continue

        for nex,ecost in lis[now]:
            if ret[nex] > ncost + ecost:
                ret[nex] = ncost + ecost
                heapq.heappush(q , (ret[nex] , nex))
    return ret

N,M,K = map(int,stdin.readline().split())

R = list(map(int,stdin.readline().split()))
for i in range(K):
    R[i] -= 1

lis = [ [] for i in range(M) ]
ABC = []

for i in range(M):

    A,B,C = map(int,stdin.readline().split())
    A -= 1
    B -= 1

    ABC.append( (A,B,C) )
    lis[A].append( (B,C) )
    lis[B].append( (A,C) )

ddic = {}
ddic[0] = Dijkstra(lis,0)

for nr in R:
    l,r,_ = ABC[nr]

    if l not in ddic:
        ddic[l] = Dijkstra(lis,l)
    if r not in ddic:
        ddic[r] = Dijkstra(lis,r)

#print (ddic)

inf = float("inf")

dp = [ [[inf,inf] for j in range(K)] for i in range(2**K) ]

# dp[bit][最後][LorR] = mincost

for i,nr in enumerate(R):

    l,r,c = ABC[nr]

    dp[2**i][i][0] = ddic[0][r] + c
    dp[2**i][i][1] = ddic[0][l] + c

def popcnt(x):
    ret = 0
    while x:
        ret += x%2
        x //= 2
    return ret

for bit in range(2**K):
    for lastNR in range(K):
        for lastSide in range(2):

            lastL,lastR,_ = ABC[R[lastNR]]
            lastV = [lastL,lastR][lastSide]
            
            if dp[bit][lastNR][lastSide] == inf:
                continue

            for newNR in range(K):
                newbit = 2**newNR | bit
                if bit == newbit:
                    continue

                newL,newR,newC = ABC[R[newNR]]

                
                dp[newbit][newNR][0] = min(dp[newbit][newNR][0] ,
                                           dp[bit][lastNR][lastSide] + ddic[lastV][newR] + newC )
                dp[newbit][newNR][1] = min(dp[newbit][newNR][1] ,
                                           dp[bit][lastNR][lastSide] + ddic[lastV][newL] + newC )

ans = inf
bit = 2**K-1
for lastNR in range(K):
    for lastSide in range(2):

        lastL,lastR,_ = ABC[R[lastNR]]
        lastV = [lastL,lastR][lastSide]

        ans = min(ans , dp[bit][lastNR][lastSide] + ddic[lastV][N-1])

print (ans)