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: Time Limit Exceeded)

def main():
  from array import array
  from collections import defaultdict
  from heapq import heappush,heappop,heapify
  A=array
  inf=2**31-1
  n,m,*t=map(int,open(0).read().split())
  edge=defaultdict(list)
  used,d={},{}
  l=[]
  for a,b,c in zip(t[::3],t[1::3],t[2::3]):
    if b<a:a,b=b,a
    x,y=a+n*c,b+n*c
    if b==n:
      l.append(y)
      d[y]=inf
    edge[a].append(A('L',[1,x]))
    edge[b].append(A('L',[1,y]))
    edge[x].extend([A('L',[0,a]),A('L',[0,y])])
    edge[y].extend([A('L',[0,b]),A('L',[0,x])])
    used[a]=used[b]=used[x]=used[y]=True
  used[1]=False
  edgelist=edge[1]
  heapify(edgelist)
  while edgelist:
    m=heappop(edgelist)
    if not used[m[1]]:
      continue
    d[m[1]]=m[0]
    used[m[1]]=False
    for e in edge[m[1]]:
      if used[e[1]]:
        heappush(edgelist,A('L',[e[0]+m[0],e[1]]))
  a=min([d[i]for i in l]or[inf])
  print([a,-1][a==inf])
if __name__=='__main__':
  main()