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
readline = sys.stdin.readline

from heapq import heappop as hpp, heappush as hp
inf = 10**9+7
def dijkstra(N, s, Edge):    
    dist = [inf] * N
    dist[s] = 0
    Q = [(0, s)]
    while Q:
        dn, vn = hpp(Q)
        if dn > dist[vn]:
            continue
        for df, vf in Edge[vn]:
            if dn + df < dist[vf]:
                dist[vf] = dn + df
                hp(Q, (dn + df,vf))
    return dist

def compress(L):
    L2 = list(set(L))
    L2.sort()
    C = {v : k for k, v in enumerate(L2)}
    return L2, C


N, M = map(int, readline().split())

Edge = []
VS = set(range(N))
geta = 20
Edge = [tuple(map(int, readline().split())) for _ in range(M)]
for a, b, c in Edge:
    a -= 1
    b -= 1
    VS |= {(c<<geta) | a, (c<<geta) | b}
   
_, Cv = compress(list(VS))

Vnum = len(Cv)

nEdge = [[] for _ in range(Vnum)]

for a, b, c in Edge:
    a -= 1
    b -= 1
    ca = Cv[(c<<geta) | a]
    cb = Cv[(c<<geta) | b]
    nEdge[ca].append((0, cb))
    nEdge[cb].append((0, ca))
    nEdge[a].append((1, ca))
    nEdge[ca].append((0, a))
    nEdge[b].append((1, cb))
    nEdge[cb].append((0, b))

dist = dijkstra(Vnum, 0, nEdge)
if dist[N-1] >= inf:
    dist[N-1] = -1
print(dist[N-1])