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 (2.4.0) (judge verdict: Time Limit Exceeded)

from collections import defaultdict
from heapq import heappush, heappop

INF = float('inf')

# 単一始点最短経路を求める（ダイクストラ法）
def Dijkstra(adjList, vSt):
    numV = len(adjList)
    numUsed = 0
    costs = defaultdict(lambda: INF)
    costs[vSt] = 0
    PQ = []
    heappush(PQ, (0, vSt))

    while PQ:
        c, vNow = heappop(PQ)
        if c > costs[vNow]: continue

        numUsed += 1
        if numUsed == numV: break

        costNow = c
        for v2, wt in adjList[vNow]:
            c2 = costNow + wt
            if c2 < costs[v2]:
                costs[v2] = c2
                heappush(PQ, (c2, v2))

    return costs


N, M = map(int, input().split())
adjL = defaultdict(list)
for _ in range(M):
    p, q, c = map(int, input().split())
    adjL[(p, c)].append(((q, c), 0))
    adjL[(q, c)].append(((p, c), 0))
    adjL[(p, 0)].append(((p, c), 1))
    adjL[(p, c)].append(((p, 0), 1))
    adjL[(q, 0)].append(((q, c), 1))
    adjL[(q, c)].append(((q, 0), 1))

costs = Dijkstra(adjL, (1, 0))
if costs[(N, 0)] == INF:
    print(-1)
else:
    print(costs[(N, 0)] // 2)