Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in CPP):

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

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

int dist[MAX_N],N,M,id;
vector <pair<int,int> > g[MAX_N];
int dits[MAX_N];
map <pair<int,int>,int> Map;
priority_queue <pair<int,int> > Q;

void add_edge(int u,int v,int cost1,int cost2){
	g[u].push_back(make_pair(v,cost1));
	g[v].push_back(make_pair(u,cost2));
}

int main(){
	scanf("%d%d",&N,&M);
	id=N;
	for(int i=0;i<M;i++){
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c); 
		u--; v--; 
		if(Map.find(make_pair(u,c))==Map.end()){
			Map[make_pair(u,c)]=id;
			add_edge(u,id++,1,0);
		}
		else add_edge(u,Map[make_pair(u,c)],1,0);
		if(Map.find(make_pair(v,c))==Map.end()){
			Map[make_pair(v,c)]=id;
			add_edge(v,id++,1,0);
		}
		else add_edge(v,Map[make_pair(v,c)],1,0);
		add_edge(Map[make_pair(u,c)],Map[make_pair(v,c)],0,0);
	}
	for(int i=0;i<MAX_N;i++) dist[i]=INF;
	dist[0]=0;
	Q.push(make_pair(-dist[0],0));
	while(!Q.empty()){
		int val=-Q.top().first,x=Q.top().second;
		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));
			}
		}
	}
	if(dist[N-1]==INF) printf("-1\n");
	else printf("%d\n",dist[N-1]);
	return 0;
}