Problem:
Score : 600 points

Problem Statement

Snuke's town has a subway system, consisting of N stations and M railway lines. The stations are numbered 1 through N. Each line is operated by a company. Each company has an identification number.

The i-th ( 1 \leq i \leq M ) line connects station p_i and q_i bidirectionally. There is no intermediate station. This line is operated by company c_i.

You can change trains at a station where multiple lines are available.

The fare system used in this subway system is a bit strange. When a passenger only uses lines that are operated by the same company, the fare is 1 yen (the currency of Japan). Whenever a passenger changes to a line that is operated by a different company from the current line, the passenger is charged an additional fare of 1 yen. In a case where a passenger who changed from some company A's line to another company's line changes to company A's line again, the additional fare is incurred again.

Snuke is now at station 1 and wants to travel to station N by subway. Find the minimum required fare.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq 2×10^5

1 \leq p_i \leq N (1 \leq i \leq M)

1 \leq q_i \leq N (1 \leq i \leq M)

1 \leq c_i \leq 10^6 (1 \leq i \leq M)

p_i \neq q_i (1 \leq i \leq M)

Input

The input is given from Standard Input in the following format:

N M
p_1 q_1 c_1
:
p_M q_M c_M

Output

Print the minimum required fare. If it is impossible to get to station N by subway, print -1 instead.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

8 11
1 3 1
1 4 2
2 3 1
2 5 1
3 4 3
3 6 3
3 7 3
4 8 4
5 6 1
6 7 5
7 8 5

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in PyPy3 (2.4.0) (judge verdict: Runtime Error)

from collections import*
from heapq import*
def dijkstra(s):
  d=[inf]*n
  used=[True]*n
  d[s]=0
  used[s]=False
  edgelist=[]
  for e in edge[s]:
    heappush(edgelist,e)
  while len(edgelist):
    minedge=heappop(edgelist)
    if not used[minedge[1]]:
      continue
    v=minedge[1]
    d[v]=minedge[0]
    used[v]=False
    for e in edge[v]:
      if used[e[1]]:
        heappush(edgelist,(e[0]+d[v],e[1]))
  return d
inf=10**18
n,m,*t=map(int,open(0).read().split())
if m<1:
  print(-1)
  exit()
e=defaultdict(list)
s=[]
l=[]
for a,b,c in zip(t[::3],t[1::3],t[2::3]):
  if b<a:a,b=b,a
  if b==n:l.append(b+n*c)
  e[a].append((1,a+n*c))
  e[b+n*c].append((0,b))
  e[a+n*c].append((0,b+n*c))
  s.extend([a,b,a+n*c,b+n*c])
d={i:v for v,i in enumerate(sorted(set(s)))}
n=len(d)
edge=tuple([]for _ in range(n))
for i in e:
  edge[d[i]].extend([(c,d[j])for c,j in e[i]])
a=dijkstra(0)
a=min(a[d[i]]for i in l)
print([a,-1][a==inf])