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;
#define pb push_back
#define mp make_pair
#define fi first
#define se second
int n,m,dis[2000007],u,v,c,cnt;
map < pair<int,int> , int > p;
priority_queue < pair<int,int> , vector < pair<int,int> > , greater < pair<int,int> > > pr;
vector < pair<int,int> > edge[2000007];
int main(){
	ios_base::sync_with_stdio(0);
	cin>>n>>m;
	cnt=n;
	for(int i=1;i<=2000000;i++) dis[i]=1e9;
	for(int i=1;i<=m;i++){
		cin>>u>>v>>c;
		int edu=p[mp(u,c)];
		if(edu==0){
			cnt++;
			p[mp(u,c)]=cnt;
			edu=cnt;
		}
		int edv=p[mp(v,c)];
		if(edv==0){
			cnt++;
			p[mp(v,c)]=cnt;
			edv=cnt;
		}
		edge[edu].pb(mp(edv,0));
		edge[edv].pb(mp(edu,0));
		edge[u].pb(mp(edu,1));
		edge[edu].pb(mp(u,1));
		edge[v].pb(mp(edv,1));
		edge[edv].pb(mp(v,1));
	}
	pr.push(mp(0,1));
	dis[1]=0;
	while(!pr.empty()){
		int val=pr.top().fi;
		u=pr.top().se;
		pr.pop();
		if(dis[u]!=val) continue;
		for(int i=0;i<(int)edge[u].size();i++){
			v=edge[u][i].fi;
			c=edge[u][i].se;
			//cout<<v<<" "<<u<<endl;
			if(dis[v]>val+c){
				dis[v]=val+c;
				pr.push(mp(dis[v],v));
			}
		}
	}
	if(dis[n]==1e9) cout<<-1;
	else cout<<dis[n]/2;
}