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: Memory Limit Exceeded)

# -*- coding: utf-8 -*-
import bisect
import heapq
import math
import random
import sys
from collections import Counter, defaultdict, deque
from decimal import ROUND_CEILING, ROUND_HALF_UP, Decimal
from functools import lru_cache, reduce
from itertools import combinations, combinations_with_replacement, product, permutations
from operator import add, mul, sub

sys.setrecursionlimit(10000)


def read_int():
    return int(input())


def read_int_n():
    return list(map(int, input().split()))


def read_float():
    return float(input())


def read_float_n():
    return list(map(float, input().split()))


def read_str():
    return input().strip()


def read_str_n():
    return list(map(str, input().split()))


def error_print(*args):
    print(*args, file=sys.stderr)


def mt(f):
    import time

    def wrap(*args, **kwargs):
        s = time.time()
        ret = f(*args, **kwargs)
        e = time.time()

        error_print(e - s, 'sec')
        return ret

    return wrap


def bfs01(g, s):
    d = {}
    for v in g.keys():
        d[v] = sys.maxsize
    d[s] = 0

    if s not in d:
        return d
    q = deque()
    q.append((d[s], s))

    while q:
        c, u = q.popleft()
        for v in g[u]:
            alt = g[u][v] + d[u]
            if d[v] > alt:
                d[v] = alt
                if g[u][v] == 0:
                    q.appendleft((d[v], v))
                else:
                    q.append((d[v], v))
    return d


@mt
def slv(N, M, PQC):

    g = defaultdict(dict)
    NM = 30
    for p, q, c in PQC:
        cp = (c << NM) + p
        cq = (c << NM) + q
        g[cp][cq] = 0
        g[cq][cp] = 0
        g[cp][p] = 0
        g[cq][q] = 0
        g[p][cp] = 1
        g[q][cq] = 1

    d = bfs01(g, 1)
    return -1 if N not in d else d[N]


def main():
    N, M = read_int_n()
    PQC = [read_int_n() for _ in range(M)]
    print(slv(N, M, PQC))


if __name__ == '__main__':
    main()