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<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define MAXN 100000
#define MAXM 200005
#define INF 0x3f3f3f3f
#define LL long long
int n,m;
/*struct node{
	int v,w;
};*/
vector<pair<int,int> >G[MAXN*5+5];
map<int,int> no[MAXN+5];
int dist[MAXN*5+5];
bool vis[MAXN*5+5];
void sp()
{
	fill(dist,dist+MAXN*5+5,INF);
	dist[1]=0;
	vis[1]=0;
	queue<int> que;
	que.push(1);
	while(!que.empty())
	{
		int u=que.front();
		que.pop();
		for(int i=0;i<G[u].size();i++)
		{
			int v=G[u][i].first,w=G[u][i].second;
			if(dist[v]>dist[u]+w)
			{
				dist[v]=dist[u]+w;
				if(!vis[v])
				{
					vis[v]=1;
					que.push(v);
				}
			}
		}
		vis[u]=0;
	}
}
int main()
{
	//freopen("trip.in","r",stdin);
	//freopen("trip.out","w",stdout);
	scanf("%d %d",&n,&m);
	int uu,vv,c,cnt=n;
	for(int i=1;i<=m;i++)
	{
		scanf("%d %d %d",&uu,&vv,&c);
		if(!no[uu].count(c)) no[uu][c]=++cnt;
		if(!no[vv].count(c)) no[vv][c]=++cnt;
		int nu=no[uu][c],nv=no[vv][c];
		G[uu].push_back(make_pair(nu,1));
		G[nu].push_back(make_pair(uu,1));
		G[vv].push_back(make_pair(nv,1));
		G[nv].push_back(make_pair(vv,1));
		G[nu].push_back(make_pair(nv,0));
		G[nv].push_back(make_pair(nu,0));
	}
	sp();
	if(dist[n]==INF)
		dist[n]=-1;
	else dist[n]/=2;
	printf("%d\n",dist[n]);
	return 0;
}