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:
#pragma GCC optimize("O3")
#include<bits/stdc++.h>
#define rc(x) return cout<<x<<endl,0
#define pb push_back
#define in insert
#define er erase
#define fd find
#define fr first
#define sc second
typedef long long ll;
const long long llinf=LLONG_MAX;
const int inf=INT_MAX;
const int nmax=1e5+5;
const int mod=1e9+7;
using namespace std;
int n,m,x,y,c,i,nxt,cs,dist[nmax];
vector<pair<int,int> >a[nmax];
set<int>d[nmax];
priority_queue<pair<int,int> >s;
int main()
{
    //freopen("sol.in","r",stdin);
	//freopen("sol.out","w",stdout);
	ios_base::sync_with_stdio(false);cin.tie(0);cerr.tie(0);cout.tie(0);
    cin>>n>>m;
    for(i=2;i<=n;i++)dist[i]=inf;
    for(i=1;i<=m;i++)
	{
		cin>>x>>y>>c;
		a[x].pb(make_pair(y,c));
		a[y].pb(make_pair(x,c));
	}
	s.push({0,1});
	while(s.size())
	{
		nxt=s.top().sc;
		cs=-s.top().fr;
		s.pop();
		if(dist[nxt]<cs)continue;
		for(i=0;i<a[nxt].size();i++)
		{
			int tmp=dist[nxt];
			if(!d[nxt].count(a[nxt][i].sc))tmp++;
			if(tmp<dist[a[nxt][i].fr])
			{
				dist[a[nxt][i].fr]=tmp;
				d[a[nxt][i].fr].clear();
				d[a[nxt][i].fr].in(a[nxt][i].sc);
				s.push({-dist[a[nxt][i].fr],a[nxt][i].fr});
			}
			else if(tmp==dist[a[nxt][i].fr])d[a[nxt][i].fr].in(a[nxt][i].sc);
		}
	}
	if(dist[n]==inf)cout<<-1<<endl;
	else cout<<dist[n]<<endl;
	return 0;
}