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 (3.4.3) (judge verdict: Time Limit Exceeded)

from collections import defaultdict,deque
N,M=map(int,input().split())
G0=defaultdict(set)
G1=defaultdict(set)
Comps=[[] for i in range(N)]
for i in range(M):
  a,b,c=map(int,input().split())
  G0[(a-1,c-1)].add((b-1,c-1))
  G0[(a-1,c-1)].add((a-1,-1))
  G0[(b-1,c-1)].add((b-1,-1))
  G0[(b-1,c-1)].add((a-1,c-1))
  Comps[a-1].append(c-1)
  Comps[b-1].append(c-1)
for i in range(N):
  for c in Comps[i]:
    G1[(i,-1)].add((i,c))
def BFS(x):
  visited=defaultdict(bool)
  stack=deque([(x,-1,0)])
  while stack:
    a,c,m=stack.popleft()
    if visited[(a,c)]:
      continue
    if a==N-1 and c==-1:
      return m
    visited[(a,c)]=True
    for i in G0[(a,c)]:
      if not visited[i]:
        stack.appendleft((i[0],i[1],m))
    for i in G1[(a,c)]:
      if not visited[i]:
        stack.append((i[0],i[1],m+1))
  return -1

print(BFS(0))