from heapq import *
from collections import defaultdict
from sys import stdin

def readNums():
    return [int(x) for x in stdin.readline().split()]

planets, no_roads = readNums()
# if we do roads[key].append(value) and key wasn't set before, we get roads[key] == [value]
# avoids doing
# if key in roads: roads[key].append(value)
# else: roads[key] = [value]
roads = defaultdict(list)
for _ in range(no_roads):
    a, b, cost = readNums()
    roads[a-1].append((b-1, cost))
    roads[b-1].append((a-1, cost))
waittimes = []
for _ in range(planets):
    times = readNums()
    waittimes.append(set(times[1:]))

def dijkstra(planets, roads, waittimes):
    " Returns an list containing the distances of cities from source "
    # assume cities are numbered [0..cities-1]
    times = [10**10] * planets # at the beginning, mark distances as infinity
    times[0] = 0
    heap = [(0, 0)] # time as first element, planet as second
    while len(heap) > 0:
        time, planet = heappop(heap) # remove the element with smallest time
        #print time, planet
        if time > times[planet]: continue # skip, we were here faster before
        while time in waittimes[planet]: # travellers arriving
            time += 1
        for nextPlanet, cost in roads[planet]:
            if times[nextPlanet] > time + cost: # if we get there faster this way
                times[nextPlanet] = time + cost
                heappush(heap, (time + cost, nextPlanet)) # next place to the heap
    return times

times = dijkstra(planets, roads, waittimes)
if times[planets-1] < 10**10:
    print times[planets-1]
else:
    print -1