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: Accepted)

//A,Elebereth Gilthoniel mantae!
//For Temeria!
#include<iostream>
#include<algorithm>
#include<cmath>
#include<string>
#include<vector>
#include<set>
#include<cstdio>
#include<map>
#include<stack>
#include<queue>
#include<cstring> 
using namespace std;
const long long maxn=2e6+1e5;
vector<pair<long long,long long> > v[maxn];vector<long long> col[maxn];long long vis[maxn];
long long bark[maxn];vector<long long> qzr[maxn];long long ful;
long long dis[maxn];
int main()
{
	long long n,m;cin>>n>>m;
	for(long long i=0;i<m;i++)
	{
		long long a,b,c;cin>>a>>b>>c;
		v[a].push_back(make_pair(c,b));v[b].push_back(make_pair(c,a));
		col[c].push_back(a);col[c].push_back(b);
	} 
	for(long long i=1;i<=n;i++)
	{
		sort(v[i].begin(),v[i].end());
	}
	ful=1e6+5;
	for(long long asd=1;asd<=1e6+500;asd++)
	{
		if(col[asd].size()==0) continue;
		for(long long it=0;it<col[asd].size();it++)
		{
			long long  p=col[asd][it];
			if(vis[p]==asd) continue;
			queue<long long> q;vector<long long> kl;kl.clear();
			q.push(p);vis[p]=asd;kl.push_back(p);
			while(!q.empty())
			{
				if(q.size()==0) break; 
				long long t=q.front();q.pop();
				for(long long i=max(bark[t]-10,(long long)0);i<v[t].size();i++)
				{
					pair<long long,long long> u=v[t][i];
					if(vis[u.second]==asd) continue;
					if(u.first>asd)
					{
						bark[t]=i;break;
					}	
					if(u.first<asd)
					{
						continue;
						//bark[t]=i;break;
					}
					kl.push_back(u.second);q.push(u.second);vis[u.second]=asd;
				}
			}
			for(long long i=0;i<kl.size();i++)
			{
			//	cout<<kl[i]<<" ";
				long long t=kl[i];qzr[t].push_back(ful),qzr[ful].push_back(t);
			}//cout<<"\n";
			ful++;
		}
	}
	memset(dis,-1,sizeof(dis));
	priority_queue<pair<long long,long long> > q;
	q.push(make_pair(-0,1));dis[1]=0;
	while(!q.empty())
	{
		if(q.size()==0) break; 
		long long t=(q.top()).second;q.pop();	
		for(long long i=0;i<qzr[t].size();i++)
		{
			long long j=qzr[t][i];
			if(dis[j]==-1)
			{
				dis[j]=dis[t]+1;q.push(make_pair(-dis[j],j));
			}

		}
	}
	if(dis[n]==-1) dis[n]--;
	cout<<dis[n]/2;
	return 0;
}