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:
//by (111110101)2
#include<iostream>
#include<sstream>
#include<fstream>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<algorithm>
#include<functional>
#include<utility>
#include<bitset>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cstdio>
using namespace std;
int n,m,it;
map<pair<int,int>,int>mp;
vector<pair<int,int> >g[606060];
bool vis[606060];
void lj(int a,int b,int c,int d)
{
	g[a].push_back(make_pair(b,c));
	g[b].push_back(make_pair(a,d));
}
int gd(int st,int c)
{
	if(mp.count(make_pair(st,c)))
	{
		return mp[make_pair(st,c)];
	}
	int now=it;
	it++;
	lj(st,now,1,0);
	mp[make_pair(st,c)]=now;
	return now;
}
int main()
{
	cin>>n>>m;
	it=n;
	for(int i=0;i<m;i++)
	{
		int a,b,c;
		cin>>a>>b>>c;
		a--;
		b--;
		int x=gd(b,c),y=gd(a,c);
		lj(x,y,0,0);
	}
	priority_queue<pair<int,int> >q;
	for(int i=0;i<it;i++)
	{
		vis[i]=false;
	}
	q.push(make_pair(0,0));
	while(!q.empty())
	{
		pair<int,int>pp=q.top();
		q.pop();
		int ct=-pp.first,x=pp.second;
		if(x==n-1)
		{
			cout<<ct<<endl;
			return 0;
		}
		for(int i=0;i<g[x].size();i++)
		{
			if(!vis[g[x][i].first])
			{
				vis[g[x][i].first]=true;
				q.push(make_pair(-(ct+g[x][i].second),g[x][i].first));
			}
		}
	}
	cout<<-1<<endl;
	return 0;
}