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 C# (Mono 4.6.2.0) (judge verdict: Time Limit Exceeded)

using System;
using System.Collections.Generic;
using System.Linq;

struct Z2 {
	public int x, y;
	public Z2(int x, int y){
		this.x = x; this.y = y;
	}
}

public class A {
	
	
	public A(){
	}
	
	string[] Rs(){
		return Console.ReadLine().Split(' ');
	}
	public void run(){
		int MAX_N = 100008;
		int MAX_M = 200008;
		int N,M;
		
		List<Z2>[] edges= new List<Z2>[MAX_N];
		for(int i=0;i<MAX_N;i++) edges[i] = new List<Z2>();
		Dictionary<Z2, int> dis  = new Dictionary<Z2,int>(5*MAX_M);
		Queue<Z2> que =  new Queue<Z2>(2*MAX_M);
		Queue<int> que2 =  new Queue<int>(MAX_M);
		
		
		string[] str = Rs();
		N = int.Parse(str[0]); M = int.Parse(str[1]);
		for(int i=0;i<M;i++){
			int[] inp = Rs().Select( x => int.Parse(x)-1 ).ToArray();
			edges[ inp[0] ].Add( new Z2(inp[1],inp[2]) );
			edges[ inp[1] ].Add( new Z2(inp[0],inp[2]) );
		}
		
		
		
		que.Enqueue(new Z2(0, -1) ); // (position, company)
		dis.Add(new Z2(0,-1), 0);
		
		while(que.Count != 0){
			Z2 n = que.Dequeue();
			if(n.x == N-1){ 
				Console.WriteLine(dis[n]);
				return;
			}
			
			que2.Clear();
			que2.Enqueue(n.x);
			while(que2.Count!=0){
				int now = que2.Dequeue();
				foreach(Z2 e in edges[now]){
					Z2 cd = new Z2(now, n.y);
					if(e.x==N-1 && e.y==n.y){
						Console.WriteLine(dis[n]);
						return;
					}
					if(e.y == n.y){
						if( !dis.ContainsKey(e) ){
							que2.Enqueue(e.x);
							dis.Add(e, dis[cd]);
						} else if(dis[e] > dis[cd]){
							que2.Enqueue(e.x);
							dis[e] =  dis[cd];
						}
					} else {
						if( !dis.ContainsKey(e) ){
							que.Enqueue(e);
							dis.Add(e, dis[n]+1);
						} else if( dis[e] > dis[n]+1){
							que.Enqueue(e);
							dis[e] = dis[n]+1;
						}
					}
				}
			}
		} 
		Console.WriteLine("-1");
	}
}

public class Program {
	static void Main(string[] args){
		A a = new A();
		a.run();
	}
}