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, deque

n, m = map(int, input().split())
G = defaultdict(set)
U = 10**6
seen = set()
for _ in range(m):
  p, q, c = map(int, input().split())
  G[p+U*c].add((q+U*c, 0))
  G[q+U*c].add((p+U*c, 0))
  G[p+U*c].add((p-U, 0))
  G[q+U*c].add((q-U, 0))
  G[p-U].add((p+U*c, 1))
  G[q-U].add((q+U*c, 1))
que = deque()
que.append((1-U, 0))
while que:
  v, cnt = que.popleft()
  if v in seen:
    continue
  if v == n-U:
    print(cnt)
    exit()
  seen.add(v)
  for nv, cost in G[v]:
    if cost == 0:
      que.appendleft((nv, cnt+cost))
    else:
      que.append((nv, cnt+cost))
print(-1)