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:
N, M = map(int, input().split())
ES = []
D = [{} for i in range(N)]

*parent, = range(M)
def root(x):
    if x == parent[x]:
        return x
    parent[x] = y = root(parent[x])
    return y
def unite(x, y):
    px = root(x); py = root(y)
    if px < py:
        parent[py] = px
    else:
        parent[px] = py

E = {}
for i in range(M):
    p, q, c = map(int, input().split()); p -= 1; q -= 1
    ES.append((p, q, c))
    if c in D[p]:
        unite(D[p][c], i)
    else:
        D[p][c] = i
    E.setdefault(p, set()).add(c)
    if c in D[q]:
        unite(D[q][c], i)
    else:
        D[q][c] = i
    E.setdefault(q, set()).add(c)
P = {}
for i in range(M):
    p, q, c = ES[i]
    s = P.setdefault(root(i), set())
    s.add(p); s.add(q)

from collections import deque
que = deque([(0, 0, 0)])
INF = 10**18
dist = [INF]*N
dist[0] = 0
gdist = [INF]*M

while que:
    cost, v, t = que.popleft()
    if t:
        # edge
        if gdist[v] < cost or v not in P:
            continue
        for w in P[v]:
            if cost + 1 < dist[w]:
                dist[w] = cost + 1
                que.append((cost + 1, w, 0))
    else:
        # node
        if dist[v] < cost or v not in E:
            continue
        for w in D[v].values():
            w = root(w)
            if cost < gdist[w]:
                gdist[w] = cost
                que.appendleft((cost, w, 1))

print(dist[N-1] if dist[N-1] < INF else -1)