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 resolve():
    def Dijkstra(s, g):
        from collections import deque
        d = {s}
        queue = deque()
        for q in edge[s]:
            queue.append((1, q))
        while queue:
            dist, point = queue.popleft()
            if point in d:
                continue
            d.add(point)
            if point == g:
                return dist
            for nextp in edge[point]:
                if not nextp in d:
                    if not -1 in point or -1 in nextp:
                        queue.appendleft((dist, nextp))
                    else:
                        queue.append((dist + 1, 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, c)].add((q, c))
        edge[(q, c)].add((p, c))
        edge[(p, c)].add((p, -1))
        edge[(q, c)].add((q, -1))
        edge[(p, -1)].add((p, c))
        edge[(q, -1)].add((q, c))
    print(Dijkstra((0, -1), (n - 1, -1)))


if __name__ == '__main__':
    resolve()