Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in Python (3.4.3) (judge verdict: Wrong Answer)

from collections import deque
N,M = map(int, input().split())
if M == 0:
    print(-1)
    exit()
pqcs = [list(map(int, input().split())) for _ in range(M)]
c_max = max(list(map(lambda x: x[2],pqcs)))
node_num = (c_max+1)*N
Q = [deque([1]), deque([])]
distance = {1:0}

#辺へ重み付け
edges = {}
for p,q,c in pqcs:
	edge = {}
	edge[c*10**6+q] = 0
	edge[p] = 0
	edges.setdefault(c*10**6+p, {})
	edges[c*10**6+p].update(edge)
	edge = {}
	edge[c*10**6+p] = 0
	edge[q] = 0
	edges.setdefault(c*10**6+q, {})
	edges[c*10**6+q].update(edge)
	edge = {}
	edge[c*10**6+p] = 1
	edges.setdefault(p, {})
	edges[p].update(edge)
	edge = {}
	edge[c*10**6+q] = 1
	edges.setdefault(q, {})
	edges[q].update(edge)

while (Q[0] or Q[1]):
	target_min_ind = Q[0].popleft() if (Q[0]) else Q[1].popleft()
	target_node = edges[target_min_ind]

	for key, value in target_node.items():
		distance.setdefault(key, float('inf'))
		dist = distance[target_min_ind] + value
		if distance[key] > dist:
			distance[key] = dist
			Q[0].append(key) if dist == 0 else Q[1].append(key)
		if key == N:
			print(distance[N] if distance[N] != float('inf') else -1)
			exit()