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;
int n,m,num,dis[500010],Link[500010],t=0;
bool vis[500010];
map < pair< int,int >,int> mp;
struct dsa
{
	int v,nex,w;
}e[1200010];
void Insert(int xx,int yy,int zz) 
{
	e[++t].nex=Link[xx];e[t].v=yy;e[t].w=zz;Link[xx]=t;
	e[++t].nex=Link[yy];e[t].v=xx;e[t].w=zz;Link[yy]=t;
}
void Dijkstra()
{
	priority_queue < pair<int,int> > q;
	memset(dis,10,sizeof(dis));
	dis[1]=0;
	q.push(make_pair(0,1));
	while (q.size()) {
		int x=q.top().second;
		q.pop();
		if (vis[x]) continue;
		vis[x]=true;
		for (int i=Link[x];i;i=e[i].nex) 
		  if (dis[x]+e[i].w<dis[e[i].v]) {
		  	  dis[e[i].v]=dis[x]+e[i].w;
		  	  q.push(make_pair(-dis[e[i].v],e[i].v));
		  }
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	num=n;
	for (int i=1;i<=m;i++) {
		int xx,yy,zz;
		scanf("%d%d%d",&xx,&yy,&zz);
		if (mp.find(make_pair(xx,zz))==mp.end()) mp[make_pair(xx,zz)]=++num;
		if (mp.find(make_pair(yy,zz))==mp.end()) mp[make_pair(yy,zz)]=++num;
		Insert(mp[make_pair(xx,zz)],mp[make_pair(yy,zz)],0);
		Insert(mp[make_pair(xx,zz)],xx,1);
		Insert(mp[make_pair(yy,zz)],yy,1);
	}
	Dijkstra();
	if (dis[n]==dis[0]) cout<<"-1"<<endl;else cout<<dis[n]/2;
	return 0;
}