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
    from collections import deque
    input = sys.stdin.readline

    N, M = map(int, input().split())
    adj = [[] for _ in range(N+1)]
    c2v = {}
    v = N+1
    for _ in range(M):
        p, q, c = map(int, input().split())
        P = p*(10**7) + c
        Q = q*(10**7) + c
        if P not in c2v:
            c2v[P] = v
            adj[p].append((v, 1))
            adj.append([(p, 1)])
            v += 1
        if Q not in c2v:
            c2v[Q] = v
            adj[q].append((v, 1))
            adj.append([(q, 1)])
            v += 1
        adj[c2v[P]].append((c2v[Q], 0))
        adj[c2v[Q]].append((c2v[P], 0))

    que = deque()
    que.append(1)
    seen = [-1] * len(adj)
    seen[1] = 0
    while que:
        v = que.popleft()
        for u, cost in adj[v]:
            if seen[u] == -1:
                seen[u] = seen[v] + cost
                if cost:
                    que.append(u)
                else:
                    que.appendleft(u)
    print(seen[N] // 2)


if __name__ == '__main__':
    main()