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:
import sys
input = sys.stdin.readline
from collections import defaultdict

"""
(駅、会社)を頂点にグラフを持つ。頂点数O(M)。
そのまま辺を貼ると辺が多くなりすぎる。
(駅、会社) -> (駅、無属性) -> (駅、会社)
"""

L = 32
mask = (1 << L) - 1
N,M = map(int,input().split())
graph = defaultdict(list)
for _ in range(M):
    p,q,c = map(int,input().split())
    p <<= L
    q <<= L
    graph[p].append(p+c)
    graph[p+c].append(p)
    graph[q].append(q+c)
    graph[q+c].append(q)
    graph[p+c].append(q+c)
    graph[q+c].append(p+c)

INF = 10 ** 9
dist = defaultdict(lambda: INF)

start = 1 << L
goal = N << L

q = [start] # 0 が会社に属していない状態
dist[start] = 0
d = 0
while q:
    qq = []
    while q:
        x = q.pop()
        if x & mask == 0:
            for y in graph[x]:
                if dist[y] <= d + 1:
                    continue
                dist[y] = d + 1
                qq.append(y)
        else:
            for y in graph[x]:
                if dist[y] <= d:
                    continue
                dist[y] = d
                q.append(y)
    d += 1
    q = qq

answer = dist[goal]
if answer == INF:
    answer = -1
print(answer)