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<iostream>
#include<vector>
#include<queue>
#include<map>
using namespace std;
int n,m;
map<int,vector<pair<int,int> > >G[1<<17];
map<int,int>d[1<<17];
int main()
{
	cin>>n>>m;
	for(int i=0;i<m;i++)
	{
		int a,b,c;cin>>a>>b>>c;
		G[a][c].push_back(make_pair(b,c));
		G[b][c].push_back(make_pair(a,c));
		G[a][c].push_back(make_pair(a,0));
		G[b][c].push_back(make_pair(b,0));
		G[a][0].push_back(make_pair(b,c));
		G[b][0].push_back(make_pair(a,c));
	}
	priority_queue<pair<int,pair<int,int> > >P;
	d[1][0]=0;
	P.push(make_pair(0,make_pair(1,0)));
	while(!P.empty())
	{
		int dis=-P.top().first;
		int u=P.top().second.first;
		int c=P.top().second.second;
		P.pop();
		if(u==n&&c==0)
		{
			cout<<dis<<endl;
			return 0;
		}
		if(d[u][c]<dis)continue;
		for(pair<int,int>&p:G[u][c])
		{
			int v=p.first;
			int cost=dis+!c;
			int col=p.second;
			if(d[v].find(col)==d[v].end()||d[v][col]>cost)
			{
				d[v][col]=cost;
				P.push(make_pair(-cost,make_pair(v,col)));
			}
		}
	}
	cout<<-1<<endl;
}