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 resolve():
    def Dijkstra(s, g):
        import heapq
        d = {s}
        queue = []
        for q in edge[s]:
            heapq.heappush(queue, q)
        while queue:
            dist, point = heapq.heappop(queue)
            q = [point]
            while q:
                point = q.pop()
                if point in d:
                    continue
                d.add(point)
                if point == g:
                    return dist
                for nextdist, nextp in edge[point]:
                    if not nextp in d:
                        if nextdist:
                            heapq.heappush(queue, (dist + 1, nextp))
                        else:
                            q.append(nextp)
        return -1
    import sys
    input = sys.stdin.readline
    from collections import defaultdict
    n, m = map(int, input().split())
    edge = defaultdict(set)
    for _ in range(m):
        p, q, c = map(int, input().split())
        p, q = p - 1, q - 1
        edge[(p, -1)].add((1, (p, c)))
        edge[(q, -1)].add((1, (q, c)))
        edge[(p, c)].add((0, (p, -1)))
        edge[(q, c)].add((0, (q, -1)))
        edge[(p, c)].add((0, (q, c)))
        edge[(q, c)].add((0, (p, c)))
    print(Dijkstra((0, -1), (n - 1, -1)))


if __name__ == '__main__':
    resolve()