import sys
readline = sys.stdin.readline

# dp[v][S] = 今までに触った猪の集合がSでvにいるときの最小コスト

N,M,K = map(int,readline().split())
R = list(map(int,readline().split()))
Rdic = {} # 道の番号から猪の番号を導く
for i in range(K):
  Rdic[R[i] - 1] = i
  
G = [[] for i in range(N)]
for i in range(M):
  a,b,c = map(int,readline().split())
  ino = -1
  if i in Rdic: # 道の番号が含まれる
    ino = Rdic[i] # イノシシの番号に変換
  G[a - 1].append([b - 1, c, ino])
  G[b - 1].append([a - 1, c, ino])
  
INF = 1 << 60
dp = [[INF] * (1 << K) for i in range(N)]
import heapq as hq
q = []
hq.heappush(q, (0, 0, 0)) # 距離、頂点、今までに触った猪の集合
complete = (1 << K) - 1
while q:
  d,v,status = hq.heappop(q)
  if dp[v][status] != INF:
    continue
  dp[v][status] = d
  if v == N - 1 and status == complete:
    print(dp[v][status])
    break
  for child, cost, ino in G[v]:
    if ino != -1:
      if dp[child][status | (1 << ino)] != INF:
        continue
      hq.heappush(q, (d + cost, child, status | (1 << ino)))
    else:
      if dp[child][status] != INF:
        continue
      hq.heappush(q, (d + cost, child, status))