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:
/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#include<queue>
using namespace std;
const int MAXN=(int)1e5;
const int INF=(int)1e9;
int n_pnt,n_edg,cnt;
map<int,int> ID[MAXN+5];
int input[2*MAXN+5][3];
vector<pair<int,int> > lnk[5*MAXN+5];
int dist[5*MAXN+5];
bool vis[5*MAXN+5];
int main()
{
	scanf("%d%d",&n_pnt,&n_edg);
	cnt=n_pnt+1;
	for(int i=0;i<n_edg;i++)
	{
		int u,v,c;scanf("%d%d%d",&u,&v,&c);
		input[i][0]=u;input[i][1]=v;input[i][2]=c;
		if(!ID[u].count(c)) ID[u][c]=cnt++;
		if(!ID[v].count(c)) ID[v][c]=cnt++;
		lnk[ID[u][c]].push_back(make_pair(ID[v][c],0));
		lnk[ID[v][c]].push_back(make_pair(ID[u][c],0));
	}
	for(int i=1;i<=n_pnt;i++)
		for(map<int,int>::iterator it=ID[i].begin();it!=ID[i].end();it++)
		{
			int id=it->second;
			lnk[i].push_back(make_pair(id,1));
			lnk[id].push_back(make_pair(i,1));
		}
	fill(dist,dist+5*MAXN+5,INF);
	dist[1]=0;vis[1]=true;
	queue<int> que;que.push(1);
	while(!que.empty())
	{
		int u=que.front();que.pop();
		for(int i=0;i<lnk[u].size();i++)
		{
			int v=lnk[u][i].first,l=lnk[u][i].second;
			if(dist[v]>dist[u]+l)
			{
				dist[v]=dist[u]+l;
				if(!vis[v])
					vis[v]=true,que.push(v);
			}
		}
		vis[u]=false;
	}
	if(dist[n_pnt]==INF) printf("%d\n",-1);
	else printf("%d\n",dist[n_pnt]/2);
	return 0;
}