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: Memory Limit Exceeded)

# coding: utf-8
# Your code here!
import sys
readline = sys.stdin.readline
read = sys.stdin.read

n,m,*pqc = map(int, read().split())

g = {}
M = iter(pqc)

for p,q,c in zip(M,M,M):
    pc = ((p-1)<<20)+c
    qc = ((q-1)<<20)+c
    pp = (p-1)<<20
    qq = (q-1)<<20
    
    if pc not in g: g[pc] = []
    if qc not in g: g[qc] = []
    if pp not in g: g[pp] = []
    if qq not in g: g[qq] = []
    
    g[pc].append(qc)
    g[pc].append(pp)

    g[qc].append(pc)
    g[qc].append(qq)

    g[pp].append(pc)
    g[qq].append(qc)

if 0 not in g: g[0] = []

from collections import deque
q = deque([0])
res = {0:0}

mask = (1<<20)-1
INF = 1<<29
while q:
    vldv = q.popleft()
    vl,dv = vldv>>20, vldv&mask
    if res[vl] < dv: continue
    #if vl==(n-1)<<20: break

    for tl in g[vl]:
        ndv = dv + (vl&mask==0 or tl&mask==0)
        if tl not in res or res[tl] > ndv:
            res[tl] = ndv
            if vl&mask==0 or tl&mask==0:
                q.appendleft((tl<<20)+ndv)
            else:
                q.append((tl<<20)+ndv)
            

if (n-1)<<20 in res:
    print(res[(n-1)<<20]//2)
else:
    print(-1)