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;
typedef long long ll;
typedef pair<ll,ll> ii;
const int N=2e6+20;
const ll inf=2e15;
ll n,m,num;
map<ll,ll> mp[N];
vector<ii> e[N];
ll inq[N],d[N];
queue<int> q;
int getid(int u,int c)
{
	if(mp[u][c]==0)
		mp[u][c]=++num;
	return mp[u][c];
}
void SPFA(int s,int t)
{
	for(int i=1;i<=num;i++)
		d[i]=inf,inq[i]=0;
	q.push(s);
	d[s]=0,inq[s]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop(),inq[u]=0;
		for(int i=0;i<e[u].size();i++)
		{
			int v=e[u][i].first,w=e[u][i].second;
			if(d[v]>d[u]+w)
			{
				d[v]=d[u]+w;
				if(!inq[v])
					inq[v]=1,q.push(v);
			}
		}
	}
	printf("%lld\n",d[t]>=inf?-1:d[t]/2);
}
int main()
{
	while(cin>>n>>m)
	{
		for(int i=0;i<=n;i++)
			e[i].clear();
		int u,v,c;
		num=n;
		while(m--)
		{
			scanf("%d%d%d",&u,&v,&c);
			int uc=getid(u,c);
			int vc=getid(v,c);
			e[u].push_back(ii(uc,1));
			e[uc].push_back(ii(u,1));
			e[v].push_back(ii(vc,1));
			e[vc].push_back(ii(v,1));
			
			e[uc].push_back(ii(vc,0));
			e[vc].push_back(ii(uc,0));		
		}
		SPFA(1,n);
		
		for(int i=0;i<=num;i++)
			mp[i].clear();
	}
	return 0;
}