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;
const int N=2000002;
struct node{
	int to,ne;
}e[N<<1];
int h[N],tot,x[N],y[N],z[N],dis[N],n,m,fa[N],i,u,v,now,xx,Q[N],mx,cnt;
queue<int>q;
vector<int>vec[N];
void add(int x,int y){e[++tot]={y,h[x]},h[x]=tot;}
void ADD(int x,int y){add(x,y),add(y,x);}
int find(int x){return x==fa[x]?x:fa[x]=find(fa[x]);}
void uni(int x,int y){
	int rx=find(x),ry=find(y);
	if (rx!=ry) fa[rx]=ry;
}
int main(){
	scanf("%d%d",&n,&m);
	for (i=1;i<=m;i++) scanf("%d%d%d",&x[i],&y[i],&z[i]),vec[z[i]].push_back(i),mx=max(mx,z[i]);
	now=n;
	for (i=1;i<=mx;i++){
		for (auto j:vec[i]) fa[x[j]]=x[j],fa[y[j]]=y[j];
		for (auto j:vec[i]) uni(x[j],y[j]);
		cnt=0;
		for (auto j:vec[i]) Q[++cnt]=find(x[j]);
		sort(Q+1,Q+cnt+1);
		cnt=unique(Q+1,Q+cnt+1)-Q-1;
		for (auto j:vec[i]){
			xx=lower_bound(Q+1,Q+cnt+1,find(x[j]))-Q;
			ADD(x[j],now+xx),ADD(y[j],now+xx);
		}
		now+=cnt;
	}
	memset(dis,63,sizeof(dis));
	q.push(1),dis[1]=0;
	while (!q.empty()){
		u=q.front(),q.pop();
		for (i=h[u];i;i=e[i].ne)
			if (dis[v=e[i].to]>1e9) dis[v]=dis[u]+1,q.push(v);
	}
	printf("%d",dis[n]>1e9?-1:dis[n]/2);
}