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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        ArrayList<Edge>[] map = new ArrayList[n+1];
        for(int i=0;i<=n;i++) map[i] = new ArrayList<>();
        int m = sc.nextInt();
        for(int i=0;i<m;i++){
            int p = sc.nextInt();
            int q = sc.nextInt();
            int c = sc.nextInt();
            map[p].add(new Edge(q,c));
            map[q].add(new Edge(p,c));
        }
        PriorityQueue<int[]> pq = new PriorityQueue<>(10,new myComparator());
        int[] dist = new int[n+1];
        Arrays.fill(dist,Integer.MAX_VALUE/2);
        dist[1] = 0;
        // int[] = {current_position, dist[current_position],company that comes from}
        for(Edge e:map[1]) {
            pq.offer(new int[]{e.to,1,e.com});
            dist[e.to]=1;
        }
        while(pq.size()>0){
            int[] cur = pq.poll();
            if(dist[cur[0]]<cur[1]) continue;
            for(Edge e:map[cur[0]]){
                int dx = e.com==cur[2]?0:1;
                if(dist[e.to]>dist[cur[0]]+dx){
                    dist[e.to] = dist[cur[0]]+dx;
                    pq.offer(new int[]{e.to,dist[e.to],e.com});
                }
            }
        }
        int ans = dist[n]==Integer.MAX_VALUE/2?-1:dist[n];
        System.out.println(ans);
    }
    static class Edge{
        int to,com;
        public Edge(int t, int c){
            this.to = t;
            this.com = c;
        }
    }
    static class myComparator implements Comparator<int[]>{
        public int compare(int[] a, int[] b){
            return a[1]-b[1];
        }
    }
}