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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;

const int INF=1e9+10;
const int MAX_N=2e6+10;

pair<int,int> dots[MAX_N];
vector <pair<int,int> > G[MAX_N];
int dist[MAX_N],N,M,n=0,m=0;
vector <pair<int,int> > g[MAX_N];
map <pair<int,int>,int> Map;
priority_queue <pair<int,int> > Q;

int main(){
	scanf("%d%d",&N,&M);
	for(int i=0;i<M;i++){
		int u,v,c; 
		scanf("%d%d%d",&u,&v,&c);
		u--; v--;
		G[u].push_back(make_pair(v,c));
		G[v].push_back(make_pair(u,c));
	}
	for(int i=0;i<N;i++){
		for(int j=0;j<G[i].size();j++){
			int c=G[i][j].second,fr=G[i][j].first;
			if(Map.find(make_pair(i,c))==Map.end()){
				dots[n].first=i; dots[n].second=c;
				Map[make_pair(dots[n].first,dots[n].second)]=n;
				n++;
			}
		}
	}
	for(int i=0;i<n;i++){
		int u=dots[i].first,c1=dots[i].second;
		for(int j=0;j<G[u].size();j++){
			int v=G[u][j].first,c2=G[u][j].second;
			int to=Map[make_pair(v,c2)];
			if(v!=u){
				if(c1==c2){
					g[i].push_back(make_pair(to,0));
					g[to].push_back(make_pair(i,0));
					m++;
				}
				else{
					g[i].push_back(make_pair(to,1));
					g[to].push_back(make_pair(i,1));
					m++;
				}
			}
		}
	}
	for(int i=0;i<MAX_N;i++) dist[i]=INF;
	for(int i=0;i<G[0].size();i++){
		int st=G[0][i].first,c=G[0][i].second;
		dist[Map[make_pair(st,c)]]=1;
		Q.push(make_pair(-dist[Map[make_pair(st,c)]],Map[make_pair(st,c)]));	
	}
	while(!Q.empty()){
		int x=Q.top().second,val=-Q.top().first;
		Q.pop();
		if(dist[x]<val) continue;
		for(int i=0;i<g[x].size();i++){
			int to=g[x][i].first,cost=g[x][i].second;
			if(dist[to]>dist[x]+cost){
				dist[to]=dist[x]+cost;
				Q.push(make_pair(-dist[to],to));
			}
		}
	}
	int ans=INF;
	for(int i=0;i<n;i++){
		int x=dots[i].first;
		if(x==N-1){
			ans=min(ans,dist[i]);		
		}
	}
	if(ans==INF) printf("-1\n");
	else printf("%d\n",ans);
	return 0;
}