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:
# https://juppy.hatenablog.com/entry/2019/04/10/ARC061_-_E_%E3%81%99%E3%81%AC%E3%81%91%E5%90%9B%E3%81%AE%E5%9C%B0%E4%B8%8B%E9%89%84%E6%97%85%E8%A1%8C_-_Python_%E7%AB%B6%E6%8A%80%E3%83%97%E3%83%AD%E3%82%B0%E3%83%A9%E3%83%9F%E3%83%B3%E3%82%B0_Atcoder
# 参考にさせていただきました。

from collections import deque, defaultdict

chg = 10**6
n, m = map(int, input().split())
edges = defaultdict(set)
# m=0のためにdefaultdictにしておくべき
visited = defaultdict(set)
for i in range(m):
    p, q, c = map(int, input().split())
    edges[p+c*chg].add(q+c*chg)
    edges[q+c*chg].add(p+c*chg)
    edges[p].add(p+c*chg)
    edges[q].add(q+c*chg)
    edges[p+c*chg].add(p)
    edges[q+c*chg].add(q)
    visited[p] = False
    visited[q] = False
    visited[p+c*chg] = False
    visited[q+c*chg] = False

ans = float("inf")
que = deque()
# now, dist
que.append((1, 0))
# 今求めるのはendの距離だけなので、dist配列はいらない
# さらに、距離は0/1なのでheapで辺を管理する必要もなく、0なら最短確定なので先に、1ならその時点での最長なので後ろにすれば良い
while que:
    now, dist = que.popleft()
    if visited[now]:
        continue
    visited[now] = True
    if now == n:
        ans = dist
        break
    for to in edges[now]:
        # 駅→ホーム
        if now < chg and to > chg:
            que.append((to, dist+1))
        else:
            que.appendleft((to, dist))
if ans == float("inf"):
    print(-1)
else:
    print(ans)