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:
from collections import deque
import sys
input = sys.stdin.buffer.readline


def bfs():
    dp[sup_v][0] = 1
    q1 = deque([sup_v])
    q2 = deque([0])
    while q1:
        v1 = q1.popleft()
        v2 = q2.popleft()
        for tmp in graph[v1 * OFFSET + v2]:
            nxt_v1, nxt_v2 = divmod(tmp, OFFSET)
            if nxt_v1 == -1:
                cost = 1
            else:
                cost = 0
            if nxt_v1 not in dp:
                dp[nxt_v1] = {}
            if nxt_v2 not in dp[nxt_v1]:
                dp[nxt_v1][nxt_v2] = dp[v1][v2] + cost
                if cost == 0:
                    q1.appendleft(nxt_v1)
                    q2.appendleft(nxt_v2)
                else:
                    q1.append(nxt_v1)
                    q2.append(nxt_v2)
            elif dp[v1][v2] + cost < dp[nxt_v1][nxt_v2]:
                dp[nxt_v1][nxt_v2] = dp[v1][v2] + cost
                if cost == 0:
                    q1.appendleft(nxt_v1)
                    q2.appendleft(nxt_v2)
                else:
                    q1.append(nxt_v1)
                    q2.append(nxt_v2)
    
n, m = map(int, input().split())
INF = 10 ** 18
OFFSET = 10 ** 6


sup_v = -1
graph = {sup_v * OFFSET: []}
for _ in range(m):
    a, b, c = map(int, input().split())
    a -= 1
    b -= 1
    if c * OFFSET + a not in graph:
        graph[c * OFFSET + a] = []
        graph[c * OFFSET + a].append(sup_v * OFFSET + a)
    if c * OFFSET + b not in graph:
        graph[c * OFFSET + b] = []
        graph[c * OFFSET + b].append(sup_v * OFFSET + b)
    if sup_v * OFFSET + a not in graph:
        graph[sup_v * OFFSET + a] = []
    if sup_v * OFFSET + b not in graph:
        graph[sup_v * OFFSET + b] = []

    graph[c * OFFSET + a].append(c * OFFSET + b)
    graph[c * OFFSET + b].append(c * OFFSET + a)
    graph[sup_v * OFFSET + a].append(c * OFFSET + b)
    graph[sup_v * OFFSET + b].append(c * OFFSET + a)


dp = {sup_v: {}}
bfs()
ans = INF
for i in dp:
    for j in dp[i]:
        if j == n - 1:
            ans = min(dp[i][j], ans)   
if ans == INF:
    print(-1)
else:
    print(ans)