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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll>Pa;
const long long INF = 1ll<<62;
const int maxn = 2e6+10;
ll n,m,cnt;
map<ll,ll>Ma[maxn];//标记
vector<Pa>V[maxn];//存放pair关系
int vis[maxn];
ll dist[maxn];
int getid(int u,int v)//返回编号 
{
	if(Ma[u][v]==0)
		Ma[u][v]=++cnt;
	return Ma[u][v];
} 
void spfa()
{
	for(int i=0;i<=cnt;i++)
	{
		vis[i]=0;
		dist[i]=INF;
	}
	queue<int>q;
	q.push(1);
	dist[1]=0;
	vis[1]=1;
	while(!q.empty())
	{
		int u=q.front();
		q.pop();
		vis[u]=0;
		for(int i=0;i<V[u].size();i++)
		{
			int v=V[u][i].first;
			int Valu=V[u][i].second;
			if(dist[v] > dist[u]+Valu)
			{
				dist[v]=dist[u]+Valu;
				if(vis[v]==0)
				{
					vis[v]=1;
					q.push(v);
				}
			}	
		}		
	}
	if(dist[n]==INF)
		cout<<"-1"<<endl;
	else
		cout<<dist[n]/2<<endl;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n>>m;
	cnt=n;//从n+1开始记录编号
	for(int i=0;i<m;i++)
	{
		int u,v,id;
		cin>>u>>v>>id;
		int uid=getid(u,id);
		int vid=getid(v,id);
		V[u].push_back(Pa(uid,1));
		V[uid].push_back(Pa(u,1));
		
		V[v].push_back(Pa(vid,1));
		V[vid].push_back(Pa(v,1));
		
		V[vid].push_back(Pa(uid,0));
		V[uid].push_back(Pa(vid,0));
	} 
	spfa();
	return 0;
}