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: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
class Program{
	static void Main(){
		int[] s=Array.ConvertAll(Console.ReadLine().Split(),int.Parse);
		int[][] a=new int[s[1]][];
		int[] h=new int[s[0]+1];
		HashSet<int>[] hs=new HashSet<int>[s[0]+1];
		List<int[]>[] li=new List<int[]>[s[0]+1];
		for(int i=1;i<=s[0];i++){
			li[i]=new List<int[]>();
			hs[i]=new HashSet<int>();
			h[i]=int.MaxValue;
		}
		h[1]=0;
		for(int i=0;i<s[1];i++){
			a[i]=Array.ConvertAll(Console.ReadLine().Split(),int.Parse);
			li[a[i][0]].Add(new int[]{a[i][1],a[i][2]});
			li[a[i][1]].Add(new int[]{a[i][0],a[i][2]});
		}
		for(int i=0;i<li[1].Count;i++){
			hpa(1,li[1][i][0],li[1][i][1]);
			h[li[1][i][0]]=1;
		}
		while(he.Count>0&&h[s[0]]==int.MaxValue){
			if(h[he[0][1]]>=he[0][0]){
				if(hs[he[0][1]].Count==0){
					hs[he[0][1]].Add(he[0][2]);
					for(int i=0;i<li[he[0][1]].Count;i++){
						int q=li[he[0][1]][i][1]==he[0][2]?he[0][0]:he[0][0]+1;
						if(h[li[he[0][1]][i][0]]>=q&&(!hs[li[he[0][1]][i][0]].Contains(li[he[0][1]][i][1]))){
							hpa(q,li[he[0][1]][i][0],li[he[0][1]][i][1]);
							h[li[he[0][1]][i][0]]=q;
						}
					}
				}
				else if(!hs[he[0][1]].Contains(he[0][2])){
					for(int i=0;i<li[he[0][1]].Count;i++){
						if(li[he[0][1]][i][1]==he[0][2]&&h[li[he[0][1]][i][0]]>=he[0][0]&&(!hs[li[he[0][1]][i][0]].Contains(li[he[0][1]][i][1]))){
							hpa(he[0][0],li[he[0][1]][i][0],li[he[0][1]][i][1]);
							h[li[he[0][1]][i][0]]=he[0][0];
						}
					}
				}
			}
			hpr();
		}
		Console.WriteLine(h[s[0]]==int.MaxValue?-1:h[s[0]]);
	}
	static List<int[]> he=new List<int[]>();
	static void hpa(int a,int b,int c){
		he.Add(new int[]{a,b,c});
		int j=he.Count-1;
		while(he[j][0]<he[(j-1)/2][0]){//
			int v=he[j][0];he[j][0]=he[(j-1)/2][0];he[(j-1)/2][0]=v;
			v=he[j][1];he[j][1]=he[(j-1)/2][1];he[(j-1)/2][1]=v;
			v=he[j][2];he[j][2]=he[(j-1)/2][2];he[(j-1)/2][2]=v;
			j=(j-1)/2;
		}
	}
	static void hpr(){
		he[0]=he[he.Count-1];
		he.RemoveAt(he.Count-1);
		int j=0;
		while(true){
			if(j*2+2<he.Count){
				if(he[j*2+1][0]<he[j*2+2][0]){j=j*2+1;}//
				else{j=j*2+2;}
			}
			else if(j*2+1<he.Count){j=j*2+1;}
			else{break;}
			if(he[j][0]>=he[(j-1)/2][0]){break;}//
			int v=he[j][0];he[j][0]=he[(j-1)/2][0];he[(j-1)/2][0]=v;
			v=he[j][1];he[j][1]=he[(j-1)/2][1];he[(j-1)/2][1]=v;
			v=he[j][2];he[j][2]=he[(j-1)/2][2];he[(j-1)/2][2]=v;
		}
	}
}