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>
#define N 1000010
using namespace std;
pair<int,int> tmp;
map<pair<int,int>,int> ind;
vector<pair<int,int> > vt[N];
priority_queue<pair<int,int> > pq;
bool book[N];
int n,m,nd,dist[N];
int main(){
	int x,y,a,b,c,i,j;
	cin>>n>>m;
	nd=n;
	for(i=0;i<m;i++)
	{
		cin>>x>>y>>c;
		x--;
		y--;
		tmp=make_pair(x,c);
		if(ind[tmp]!=0)
		{
			a=ind[tmp];
		}
		else
		{
			a=nd;
			ind[tmp]=nd;
			nd++;
		}
		vt[x].push_back(make_pair(a,1));
		vt[a].push_back(make_pair(x,0));
		tmp=make_pair(y,c);
		if(ind[tmp]!=0)
		{
			b=ind[tmp];
		}
		else
		{
			b=nd;
			ind[tmp]=nd;
			nd++;
		}
		vt[y].push_back(make_pair(b,1));
		vt[b].push_back(make_pair(y,0));
		vt[a].push_back(make_pair(b,0));
		vt[b].push_back(make_pair(a,0));
	}
	/*for(i=0;i<nd;i++)
	{
		for(j=0;j<vt[i].size();j++)
		{
			cout<<i<<"--"<<vt[i][j].second<<"-->"<<vt[i][j].first<<endl;
		}
	}*/
	pq.push(make_pair(0,0));
	while(!pq.empty())
	{
		c=-(pq.top().first);
		x=pq.top().second;
		pq.pop();
		if(book[x])
		{
			continue;
		}
		book[x]=true;
		if(x==n-1)
		{
			cout<<c<<endl;
			return 0;
		}
		for(i=0;i<vt[x].size();i++)
		{
			pq.push(make_pair(-(c+vt[x][i].second),vt[x][i].first));
		}
	}
	cout<<-1<<endl;
	return 0;
}