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 PyPy3 (7.3.0) (judge verdict: Runtime Error)

import heapq
import sys
from collections import defaultdict
input = sys.stdin.readline
N, M = map(int, input().split())
Rails = [[] for _ in range(N)]

def constant_factory(value): return lambda: value


def solve():
    Costs = defaultdict(constant_factory(1000000000))
    Visited = defaultdict(constant_factory(False))
    for _ in range(M):
        p, q, c = map(int, input().split())
        Rails[p-1].append((q-1, c))
        Rails[q-1].append((p-1, c))

    Q = [(0, 0, -1)]
    heapq.heapify(Q)
    while Q:
        currentCost, currentNode, currentCompany = heapq.heappop(Q)
        if not Visited[(currentNode, currentCompany)]:
            Visited[(currentNode, currentCompany)] = True
            if currentCompany == -1:
                for nextNode, nextCompany in Rails[currentNode]:
                    if not Visited[(nextNode, nextCompany)]:
                        if Costs[(nextNode, nextCompany)] > currentCost + 1:
                            Costs[(nextNode, nextCompany)] = currentCost + 1
                            heapq.heappush(Q, (currentCost + 1, nextNode, nextCompany))
            else:
                if not Visited[(currentNode, -1)]:
                    if Costs[(currentNode, -1)] > currentCost:
                        Costs[(currentNode, -1)] = currentCost
                        heapq.heappush(Q, (currentCost, currentNode, -1))
                for nextNode, nextCompany in Rails[currentNode]:
                    if nextCompany == currentCompany and not Visited[(nextNode, nextCompany)]:
                        if Costs[(nextNode, nextCompany)] > currentCost:
                            Costs[(nextNode, nextCompany)] = currentCost
                            heapq.heappush(Q, (currentCost, nextNode, nextCompany))
    minCost = Costs[(N-1, -1)]
    print(minCost if minCost < 1000000000 else -1)
	return 0
  
if __name__ == "__main__":
    solve()