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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<queue>
using namespace std;
int n,m;
vector<pair<pair<int,int>,int> >E;
vector<int>G[1<<17];
int d[2<<17];
main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		int a,b,c;cin>>a>>b>>c;
		E.push_back(make_pair(make_pair(a,b),c));
		G[a].push_back(i);
		G[b].push_back(i);
	}
	for(int i=0;i<m;i++)d[i]=1e9;
	priority_queue<pair<int,pair<int,int> > >P;
	for(int i=0;i<G[1].size();i++)
	{
		int id=G[1][i];
		d[id]=1;
		P.push(make_pair(-1,make_pair(G[1][i],E[id].first.first+E[id].first.second-1)));
	}
	while(!P.empty())
	{
		int cost=-P.top().first;
		int id=P.top().second.second;
		int u=P.top().second.second;
		P.pop();
		if(d[id]<cost)continue;
		int c=E[id].second;
		if(u==n)
		{
			cout<<cost<<endl;
			return 0;
		}
		for(int i=0;i<G[u].size();i++)
		{
			int idx=G[u][i];
			int col=E[idx].second;
			int costt=cost+(c!=col);
			if(d[idx]>costt)
			{
				d[idx]=costt;
				P.push(make_pair(-costt,make_pair(idx,E[idx].first.first+E[idx].first.second-u)));
			}
		}
	}
	cout<<-1<<endl;
}