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 Max_N=1e5,Max_M=2*1e5,Max_C=1e6;
int N,M,cnt=0,Num=0,lin[Max_N*5+10]={},dis[Max_N*5+10]={},Vis[Max_N*5+10]={};
map<int,map<int,int> > vis;
struct kk{
	int Id,Next,v;
} e[Max_M*6+10]={};
priority_queue< pair<int,int> > Q;
inline void insert(int x,int y,int z){
	e[++Num].Next=lin[x]; lin[x]=Num; e[Num].Id=y; e[Num].v=z;
}
inline void Dij(){
	memset(dis,10,sizeof(dis));
	Q.push(make_pair(0,1)); dis[1]=0;
	for(;Q.size();){
		int x=Q.top().second; Q.pop();
		if(Vis[x]) continue; ++Vis[x];
		for(int i=lin[x];i;i=e[i].Next)
		 if(dis[x]+e[i].v<dis[e[i].Id]){
		 	dis[e[i].Id]=dis[x]+e[i].v;
		 	Q.push(make_pair(-dis[e[i].Id],e[i].Id));
		 }
	}
	if(dis[N]==dis[0]) puts("-1");
	else printf("%d\n",dis[N]/2);
}
int main(){
	scanf("%d%d",&N,&M);
	for(int i=1;i<=M;i++){
		int p,q,c,x=0,y=0; scanf("%d%d%d",&p,&q,&c);
		if(!vis[p][c]) vis[p][c]=++cnt; x=vis[p][c];
		if(!vis[q][c]) vis[q][c]=++cnt; y=vis[q][c];
		insert(p,N+x,1); insert(N+x,N+y,0); insert(N+y,q,1);
		insert(N+x,p,1); insert(N+y,N+x,0); insert(q,N+y,1);
	}
	Dij();
	return 0;
}