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;
int dis[2000005],f[100005],w[100005],qu[5000005],cur;
vector<int>g[1000005];
vector<pair<int,int>>edge[1000005],a;
int get_f(int x)
{
	if(x==f[x])
		return x;
	return f[x]=get_f(f[x]);
}
void merge(int x,int y)
{
	if(w[x]!=cur)
	{
		w[x]=cur;
		a.push_back(make_pair(0,x));
		f[x]=x;
	}
	if(w[y]!=cur)
	{
		w[y]=cur;
		a.push_back(make_pair(0,y));
		f[y]=y;
	}
	f[get_f(x)]=get_f(y);
}
int main()
{
	int n,m,p,q,c;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&p,&q,&c);
		edge[c].push_back(make_pair(p,q));
	}
	int sum=n;
	for(int i=1;i<=1000000;i++)
	{
		if(edge[i].empty())
			continue;
		cur++;
		a.clear();
		for(int j=0;j<edge[i].size();j++)
		{
			int x=edge[i][j].first,y=edge[i][j].second;
			merge(x,y);
		}
		for(int j=0;j<a.size();j++)
			a[j].first=get_f(a[j].second);
		sort(a.begin(),a.end());
		for(int j=0,k=0;j<a.size();j=k)
		{
			sum++;
			while(k<a.size()&&a[j].first==a[k].first)
			{
				g[sum].push_back(a[k].second);
				g[a[k++].second].push_back(sum);
			}
		}
	}
	for(int i=0;i<=sum;i++)
		dis[i]=-2;
	dis[1]=0;
	qu[0]=1;
	int b=0,e=1;
	while(b<e)
	{
		int k=qu[b];
		b++;
		for(int i=0;i<g[k].size();i++)
		{
			int v=g[k][i];
			if(dis[v]==-2)
			{
				dis[v]=dis[k]+1;
				qu[e]=v;
				e++;
			}
		}
	}
	printf("%d\n",dis[n]/2);	
	return 0;
}