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;

#define FOR(i,a,b) for(int i=a;i<b;i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back

using uint=unsigned int;
using ll=long long;
using pii=pair<int,int>;
using vi=vector<int>;
using ld=long double;

int read(){
	int i;
	scanf("%d",&i);
	return i;
}

const int Nmax=114514;
const int inf=8101919;
map<int,vi> graph[Nmax];
int cost[Nmax];
bool visnow[Nmax];
void dfs(int v,int c,int belong,vi& dst,vi& trav){
	cost[v]=c;
	visnow[v]=true;
	dst.PB(v);
	trav.PB(v);
	for(auto to:graph[v][belong])
		if(cost[to]>=c&&!visnow[to])
			dfs(to,c,belong,dst,trav);
}

int main(){
	int n=read(),m=read();
	REP(i,m){
		int p=read()-1,q=read()-1,c=read()-1;
		graph[p][c].PB(q);
		graph[q][c].PB(p);
	}
	fill(cost,cost+n,inf);
	cost[0]=0;
	vi cur,trav,nx;
	cur.PB(0);
	int w=0;
	while(cost[n-1]==inf&&!cur.empty()){
		w++;
		nx.clear();
		for(auto v:cur)
			for(auto kv:graph[v])
				for(auto to:kv.second)
					if(cost[to]>=w){
						trav.clear();
						dfs(v,w,kv.first,nx,trav);
						for(auto i:trav)
							visnow[i]=false;
					}
		swap(cur,nx);
	}
	if(cost[n-1]==inf)
		cout<<-1<<endl;
	else
		cout<<cost[n-1]<<endl;
}