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<int,int> PII;
const int maxn=1e6+7;
struct Edge{
	int to,w;
};
map<int,set<int> > col[200005];
vector<Edge> G[maxn];
map<PII,int> num;
int curn=0;
inline int getid(int u,int col)
{
	if(!num.count(make_pair(u,col)))
		num[make_pair(u,col)]=curn++;
	return num[make_pair(u,col)];
}
int n,m;
struct Heap{
	int u,dis;
	bool operator<(const Heap& a)const{
		return dis>a.dis;
	}
};
int d[maxn];
bool vis[maxn];
int dij()
{
	memset(d,0x3f,sizeof(d));
	d[1]=0;
	priority_queue<Heap> q;
	q.push({1,0});
	while(!q.empty())
	{
		auto u=q.top();
		q.pop();
		if(vis[u.u]) continue;
		vis[u.u]=1;
		for(auto e:G[u.u])
		{
			int v=e.to;
			if(d[v]>d[u.u]+e.w)
			{
				d[v]=d[u.u]+e.w;
				q.push({v,d[v]});
			}
		}
	}
	return d[n];
}
int main()
{
	scanf("%d%d",&n,&m);
	curn=n+1;
	for(int i=0,u,v,c;i<m;i++)
	{
		scanf("%d%d%d",&u,&v,&c);
		col[u][c].insert(v);
		col[v][c].insert(u);
	}
	for(int i=1;i<=n;i++)
	{
		for(auto u:col[i])
		{
			int cc=u.first;
			for(auto v:u.second)
			{
				G[getid(i,cc)].push_back({getid(v,cc),0});
				G[getid(v,cc)].push_back({getid(i,cc),0});
			}
			G[i].push_back(Edge{getid(i,cc),1});
			G[getid(i,cc)].push_back(Edge{i,0});
		}
	}
	int ans=dij();
	if(ans==0x3f3f3f3f) ans=-1;
	printf("%d\n",ans);
}