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
#include<iostream>
#include<set>
#include<cstring>
#include<queue>
using namespace std;
struct edge{int to,color;};
vector<edge> w[100010];
int n,m,d[100010];
set<int> s[100010];
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>n>>m;
	while(m--)
	{
		int a,b,c;
		cin>>a>>b>>c;
		w[a].push_back({b,c});
		w[b].push_back({a,c});
	}
	memset(d,0x3f,sizeof d);
	d[1]=0;
	priority_queue<pair<int,int>> q;
	q.push({-d[1],1});
	while(!q.empty())
	{
		int now=q.top().second,cost=-q.top().first;q.pop();
		if(d[now]<cost) continue;
		for(edge j:w[now])
		{
			int dis=d[now];
			if(!s[now].count(j.color))dis++;
			if(dis<d[j.to])
			{
				d[j.to]=dis;
				s[j.to].clear();
				s[j.to].insert(j.color);
				q.push({-d[j.to],j.to});
			}
			else if(dis==d[j.to])s[j.to].insert(j.color);
		}
	}
	if(d[n]>4e5)cout<<-1<<endl;
	else cout<<d[n]<<endl;
	return 0;
}
