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: Accepted)

import sys
from collections import deque, defaultdict


def bfs01(s, t, links):
    q = deque([(0, s, -1)])  # cost, station, last-company
    visited = set()
    while q:
        d, v, e = q.popleft()
        if v == t:
            return d
        if (v, e) in visited:
            continue
        visited.add((v, e))
        if e == 0:
            q.extend((d + 1, u, c) for c in links[v] for u in links[v][c] if (u, c) not in visited)
        else:
            q.extendleft((d, u, e) for u in links[v][e] if (u, e) not in visited)
            if (v, 0) not in visited:
                q.appendleft((d, v, 0))

    return -1


readline = sys.stdin.buffer.readline
read = sys.stdin.buffer.read

n, m = map(int, readline().split())
links = [defaultdict(set) for _ in range(n)]
pqc = list(map(int, read().split()))
for p, q, c in zip(pqc[0::3], pqc[1::3], pqc[2::3]):
    p -= 1
    q -= 1
    links[p][c].add(q)
    links[q][c].add(p)

print(bfs01(0, n - 1, links))