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<cstdio>
#include<cstdlib>
#include<cstring>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
vector<pair<int,int> > e[101010];
map <pair<int,int>,vector<int> > mp;
queue <int> q; 
int dis[101010];
void work(int p,int c)
{
	vector<int> v=mp[make_pair(p,c)];
	mp[make_pair(p,c)].clear();
	for(int i=0;i<v.size();i++)
		if(dis[v[i]]==-1)
		{
			q.push(v[i]);
			dis[v[i]]=dis[p];
			work(v[i],c);
		}
		else if(dis[v[i]]==dis[p])
			work(v[i],c);
}
int main()
{
	int n,m,x,y,c,p;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&x,&y,&c);
		mp[make_pair(x,c)].push_back(y);
		mp[make_pair(y,c)].push_back(x);
		e[x].push_back(make_pair(y,c));
		e[y].push_back(make_pair(x,c));
	}
	memset(dis,-1,sizeof(dis));
	dis[1]=0;
	q.push(1);
	while(!q.empty())
	{
		p=q.front();
		q.pop();
		for(int i=0;i<e[p].size();i++)
			if(dis[e[p][i].first]==-1)
			{
				q.push(e[p][i].first);
				dis[e[p][i].first]=dis[p]+1;
				work(e[p][i].first,e[p][i].second);
			}
			else if(dis[e[p][i].first]==dis[p]+1)
				work(e[p][i].first,e[p][i].second);
	}
	printf("%d\n",dis[n]);
	return 0;
}