Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in PYTHON3):

Solution:
#!/usr/bin/env python3
import sys
input = sys.stdin.readline
from collections import deque
INF = 10**9

n, m = map(int, input().split())
pqc = []

seen = set()
for i in range(n):
    seen.add((i, 0))
for _ in range(m):
    p, q, c = map(int, input().split())
    p -= 1; q -= 1
    pqc.append((p, q, c))
    seen.add((p, c))
    seen.add((q, c))
comp = dict()
for i, node in enumerate(seen):
    comp[node] = i

edge = [[] for _ in range(len(comp))]
for key in comp.keys():
    v, c = key
    if c != 0:
        frm = comp[(v, c)]
        too = comp[(v, 0)]
        edge[frm].append((too, 0))
        edge[too].append((frm, 1))

for p, q, c in pqc:
    frm = comp[(p, c)]
    too = comp[(q, c)]
    edge[frm].append((too, 0))
    edge[too].append((frm, 0))

class BFS:
    def __init__(self, adj):
        self.adj = adj
        self.dist = [INF] * len(adj)
        self.q = deque()

    def calc(self, start):
        self.dist[start] = 0
        self.q.append((0, start))
        while len(self.q) != 0:
            prov_cost, src = self.q.popleft()
            if self.dist[src] < prov_cost:
                continue
            for dest, cost in self.adj[src]:
                if self.dist[dest] > self.dist[src] + cost:
                    self.dist[dest] = self.dist[src] + cost
                    if cost == 1:
                        self.q.append((self.dist[dest], dest))
                    else:
                        self.q.appendleft((self.dist[dest], dest))
        return self.dist

bfs = BFS(edge)
bfs.calc(comp[(0, 0)])
ans = bfs.dist[comp[(n-1, 0)]]
if ans == INF:
    print(-1)
else:
    print(ans)