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
def main():
  N,M=map(int,input().split())

  if M == 0:
    print(-1)
    exit()

  INF = 10**6
  pqc = [tuple(map(lambda x:int(x)-1,input().split())) for _ in range(M)]

  to1d={v:{} for v in range(N)}
  count=2
  for p,q,c in pqc:
    if not to1d[p].get(c):
      to1d[p][c] = count
      count += 1
    if not to1d[q].get(c):
      to1d[q][c] = count
      count += 1

  G = [{} for _ in range(N+count)]
  for p,q,c in pqc:
    v1,v2 = to1d[p][c],to1d[q][c]
    G[v1][v2] = G[v2][v1] = 0

  for i in range(N):
    if len(to1d[i])<=1: continue
    for c in to1d[i].keys():
      v = to1d[i][c]
      G[v][count] = 1
      G[count][v] = 0
    count += 1

  def dijkstra(start = 0, goal = 1):
      from heapq import heappop, heappush
      NN = len(G)
      d = [INF]*NN
      d[start] = 0
      que = []
      heappush(que, start)
      while que:
          p = divmod(heappop(que),NN)
          v = p[1]
          if d[v] < p[0]: continue
          for u in G[v].keys():
              if d[u] > d[v] + G[v][u]:
                  d[u] = d[v] + G[v][u]
                  heappush(que, d[u]*NN+u)
          if v == goal: return d[goal]
      return d[goal]

  for c in to1d[0].keys():
    v = to1d[0][c]
    G[0][v] = 1
  for c in to1d[N-1].keys():
    v = to1d[N-1][c]
    G[v][1] = 0

  ans = dijkstra(0,1)
  print(ans if ans < INF else -1)
  
if __name__=="__main__":
  main()