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 Python (3.8.2) (judge verdict: Wrong Answer)

from collections import defaultdict 

N,M=map(int,input().split())

if M == 0:
  print(-1)
  exit()

INF = 10**20
pqc = [tuple(map(lambda x:int(x)-1,input().split())) for _ in range(M)]

to1d={v:{} for v in range(N)}
count=2
for p,q,c in pqc:
  if not to1d[p].get(c):
    to1d[p][c] = count
    count += 1
  if not to1d[q].get(c):
    to1d[q][c] = count
    count += 1

G = [{} for _ in range(count)]
for p,q,c in pqc:
  v1,v2 = to1d[p][c],to1d[q][c]
  G[v1][v2] = G[v2][v1] = 0
  
for v in range(N):
  if len(to1d[v])>1:
    tmp = list(to1d[v].keys())
    for i in range(len(tmp)-1):
      for j in range(i+1,len(tmp)):
        v1,v2 = to1d[v][tmp[i]],to1d[v][tmp[j]]
        G[v1][v2] = G[v2][v1] = 1
        
for c in to1d[0].keys():
  v = to1d[0][c]
  G[0][v] = 1
for c in to1d[N-1].keys():
  v = to1d[N-1][c]
  G[v][1] = 0
  
def bfs01(s):
    from collections import deque
    seen = [0]*len(G)
    d = [INF]*len(G)
    d[s] = 0
    todo = deque()
    seen[s]=1
    todo.append(s)
    while todo:
      a = todo.popleft()
      for b in G[a].keys():
        if seen[b] == 0:
          seen[b] = 1
          if G[a][b]==0:
            todo.appendleft(b)
            d[b] = d[a]
          else:
            todo.append(b)
            d[b] = d[a] + 1
    return d
  
d = bfs01(0)
ans = d[1]
print(ans if ans < INF else -1)