Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in PYTHON3):

Solution:
def main():
    import sys
    read = sys.stdin.buffer.read
    readline = sys.stdin.buffer.readline
    readlines = sys.stdin.buffer.readlines
    sys.setrecursionlimit(10 ** 7)

    from collections import deque, defaultdict

    n, m = map(int, readline().split())
    graph = [defaultdict(set) for _ in range(n + 1)]
    for _ in range(m):
        p, q, c = map(int, readline().split())
        graph[p][c].add(q)
        graph[q][c].add(p)
    q = deque([(1, 0, -1)])
    vis = set()
    while q:
        now, v, c = q.popleft()
        if now == n:
            print(v)
            exit()
        if (now, c) in vis:
            continue
        vis.add((now, c))
        if c == 0:
            for cc in graph[now]:
                for next in graph[now][cc]:
                    if (next, cc) not in vis:
                        q.append((next, v + 1, cc))
        else:
            for next in graph[now][c]:
                if (next, c) not in vis:
                    q.appendleft((next, v, c))
            if (now, 0) not in vis:
                q.appendleft((now, v, 0))
    print(-1)


if __name__ == '__main__':
    main()