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.*;

public class Main {
    private static final int INF = 1145141919;

    class City {
        int kind = 0;
        int minCost = INF;
        boolean visit = false;
        List<Road> child = new ArrayList<>();
    }

    class Road {
        int src, dst, kind;

        public Road(int src, int dst, int kind) {
            this.src = src;
            this.dst = dst;
            this.kind = kind;
        }
    }

    public void main(Scanner sc) {
        int n = sc.nextInt();
        int m = sc.nextInt();
        City c[] = new City[n];
        for (int i = 0; i < n; i++) {
            c[i] = new City();
        }
        for (int i = 0; i < m; i++) {
            int p = sc.nextInt() - 1;
            int q = sc.nextInt() - 1;
            int kind = sc.nextInt();
            c[p].child.add(new Road(p, q, kind));
            c[q].child.add(new Road(q, p, kind));
        }

        c[0].minCost = 0;
        Queue<City> queue = new PriorityQueue<>((o1, o2) -> o1.minCost - o2.minCost);
        queue.add(c[0]);
        while (!queue.isEmpty()) {
            City city = queue.poll();
            if (city.visit) {
                continue;
            }

            city.visit = true;
            for (Road r : city.child) {
                c[r.dst].minCost =
                    Math.min(c[r.dst].minCost, city.minCost + (city.kind == r.kind ? 0 : 1));
                c[r.dst].kind = r.kind;
                queue.add(c[r.dst]);
            }
        }

        int ans = c[n - 1].minCost;
        System.out.println(ans == INF ? -1 : ans);
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        new Main().main(sc);
        sc.close();
    }
}