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;
queue <int > Q;
const int N=300005,E=800005;
int dist[N];
int n,m,id;
#define pb push_back
vector <int > v[N];
struct edge{
	int u,v,w;
}e[E];
inline bool cmp(edge a,edge b){
	return a.w<b.w;
}
int fa[N],val[N],to[N];
inline int getfa(int x){
	return fa[x]==x?x:fa[x]=getfa(fa[x]);
}
int main (){
	scanf ("%d%d",&n,&m);id=n;
	for (int i=1;i<=m;i++)
		scanf ("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
	for (int i=1;i<=n;i++) fa[i]=i;
	sort(e+1,e+m+1,cmp);
	for (int i=1;i<=m;){
		int j=i;
		while (e[j].w==e[i].w) ++j;--j;
		for (int t=i;t<=j;t++)	
			if (getfa(e[t].u)!=getfa(e[t].v))
				fa[getfa(e[t].u)]=getfa(e[t].v);
		int cnt=0;
		for (int t=i;t<=j;t++)
			val[++cnt]=getfa(e[t].u),val[++cnt]=getfa(e[t].v);
		sort(val+1,val+cnt+1);to[val[1]]=++id;
		for (int t=2;t<=cnt;t++)
			if (val[t]!=val[t-1]) to[val[t]]=++id;
		for (int t=i;t<=j;t++){
			int p=e[t].u;v[p].pb(to[fa[p]]),v[to[fa[p]]].pb(p);
			p=e[t].v;v[p].pb(to[fa[p]]),v[to[fa[p]]].pb(p);
		}
		for (int t=i;t<=j;t++)
			fa[e[t].u]=e[t].u,fa[e[t].v]=e[t].v;
		i=j+1;
	}
	Q.push(1),dist[1]=1;
	while (!Q.empty()){
		int x=Q.front(),s=v[x].size();Q.pop();
		for (int t=0;t<s;t++)
			if (!dist[v[x][t]])
				Q.push(v[x][t]),dist[v[x][t]]=dist[x]+1;
	}
	if (!dist[n]) puts("-1");
	else printf ("%d",(dist[n]-1)/2);
	return 0;
}