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:
from collections import deque
N, M = map(int, input().split())
F_C_of_S = [{} for i in range(N)]
#Companies of Station
C_of_S = {}

#def Union-find
parent = [-1]*M
def root(x):
	if (parent[x] < 0): return x
	else:
		parent[x] = y = root(parent[x])
		return y

def unite(x, y):
	px = root(x)
	py = root(y)
	if (px == py): return False
	if (parent[px] > parent[py]):
		px = root(y)
		py = root(x)
	parent[px] += parent[py]
	parent[py] = px
	return True


#Union-find
pqcs = []
for i in range(M):
	p, q, c = map(int, input().split())
	p -= 1; q -= 1;
	pqcs += [(p, q, c)]
	
	#The first
	if not c in F_C_of_S[p]: F_C_of_S[p][c] = i
	#else
	else: unite(F_C_of_S[p][c], i)
	C_of_S.setdefault(p, set()).add(c)
	
	#The first
	if not (c in F_C_of_S[q]):F_C_of_S[q][c] = i
	#else
	else: unite(F_C_of_S[q][c], i)
	C_of_S.setdefault(q, set()).add(c)

#stations of company
S_of_C = {}
for i in range(M):
	p, q, c = pqcs[i]
	c_set = S_of_C.setdefault(root(i), set())
	c_set.add(p)
	c_set.add(q)


Q = deque([(0, 0, 0)])
#cost to go to the stations from station_0
dist = [float('inf')]*N
dist[0] = 0

gdist = [float('inf')]*M

while Q:
	cost, v, flag = Q.popleft()
	
	if not (flag):#(flag == 0) then gdist_update
		if (dist[v] < cost) or (v not in C_of_S):
			continue
		for l in F_C_of_S[v].values():
			l = root(l)
			if (cost < gdist[l]):
				gdist[l] = cost
				Q.appendleft((cost, l, 1))
	
	else:#(flag == 1) then dist_update
		if (gdist[v] < cost) or (v not in S_of_C):
			continue
		for s in S_of_C[v]:
			if (cost+1 < dist[s]):
				dist[s] = cost+1
				Q.append((cost+1, s, 0))

print(dist[N - 1] if dist[N - 1] < float('inf') else -1)