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
import sys

int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def MI1(): return map(int1, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def SI(): return sys.stdin.readline()[:-1]

from collections import deque

def solve():
    dist = [inf] * len(to)
    q = deque()
    q.append((0, 0))
    dist[0] = 0
    while q:
        d, i = q.popleft()
        if d > dist[i]: continue
        if i < n:
            for j in to[i]:
                if dist[j] <= d + 1: continue
                dist[j] = d + 1
                q.append((d + 1, j))
        else:
            for j in to[i]:
                if j == n - 1:
                    print(dist[i])
                    exit()
                if dist[j] <= d: continue
                dist[j] = d
                q.appendleft((d, j))
    print(-1)

inf=10**9
n,m=MI()
to=[]
uctoi={}

for u in range(n):
    to.append([])
    uctoi[u,0]=u

for _ in range(m):
    u,v,c=MI()
    u,v=u-1,v-1
    if (u,c) not in uctoi:
        i=uctoi[u,c]=len(to)
        to.append([])
        to[i].append(u)
        to[u].append(i)
    else:i=uctoi[u,c]
    if (v,c) not in uctoi:
        j=uctoi[v,c]=len(to)
        to.append([])
        to[j].append(v)
        to[v].append(j)
    else:j=uctoi[v,c]
    to[i].append(j)
    to[j].append(i)
#print(to)
#print(fin)

solve()
