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)

#include<bits/stdc++.h>
#define pb push_back
#define fi first
#define se second

using namespace std;
typedef pair<int,int> pi;

const int N=500005,M=200005;

int n,m,k,dist[N];
bool vi[N];
int fr[M],to[M],cl[M];
vector<int> e0[N],e1[N];
map<pi,int> dat;
queue<int> q0,q1;

int makenode(int u,int c){
	pi a={u,c};
	if(!dat.count(a)){
		dat[a]=++k,e0[u].pb(k),e1[k].pb(u);
		return k;
	}
	return dat[a];	
}

void expand(int u){
	if(vi[u])return;
	vi[u]=true;
	for(int v:e0[u])if(dist[v]==-1)dist[v]=dist[u],q0.push(v);
	for(int v:e1[u])if(dist[v]==-1)dist[v]=dist[u]+1,q1.push(v);
}

int main(){
	scanf("%d%d",&n,&m);k=n;
	for(int i=1;i<=m;i++)scanf("%d%d%d",&fr[i],&to[i],&cl[i]);
	for(int i=1;i<=m;i++){
		int u=fr[i],v=to[i],c=cl[i];
		int s=makenode(u,c),t=makenode(v,c);
		e0[s].pb(t),e0[t].pb(s);
	}
	memset(dist,-1,sizeof(dist));
	q0.push(1);dist[1]=0;
	while((q0.size()||q1.size())&&dist[n]==-1){
		if(q0.size())expand(q0.front()),q0.pop();
		else expand(q1.front()),q1.pop();
	}
	printf("%d\n",dist[n]);
	return 0;
}