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<cstdio>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;
map<pair<int,int>,int >mp;
struct Edge{
	int to,nxt,weigh;
}edge[2000100];
queue<int>qu1;
int cnt,head[2000100],dis[2000100],times[2000100],Find[2000100];
int n,m,idx,x,y,z;
void addedge(int x,int y,int z){
	edge[++cnt].to=y;
	edge[cnt].nxt=head[x];
	head[x]=cnt;
	edge[cnt].weigh=z;
}
void SPFA(){
	memset(dis,0x3f,sizeof(dis));
	memset(times,0,sizeof(times));
	memset(Find,false,sizeof(Find));
	times[1]=1;
	dis[1]=0;
	Find[1]=true;
	qu1.push(1);
	while(!qu1.empty()){
		int now=qu1.front();
		for(int i=head[now];i;i=edge[i].nxt){
			int upup=edge[i].to;
			if(dis[now]+edge[i].weigh<dis[upup]){
				dis[upup]=dis[now]+edge[i].weigh;
				if(!Find[upup]){
					qu1.push(upup);
					Find[upup]=true;
					times[upup]++;
					if(times[upup]>n) return;				
				}
			}
		}
		qu1.pop();
		Find[now]=false; 
	}
}
int get(int x,int y){
	if(mp[make_pair(x,y)]) return mp[make_pair(x,y)];
	else return mp[make_pair(x,y)]=++idx;
}
int main(){
    scanf("%d%d",&n,&m);
    idx=n;
    for(int i=1;i<=m;i++){
    	scanf("%d%d%d",&x,&y,&z);
    	int id1=get(x,z),id2=get(y,z);
    	addedge(id1,id2,0),addedge(id2,id1,0);
    	addedge(x,id1,1),addedge(id1,x,1);
    	addedge(y,id2,1),addedge(id2,y,1);
	}
	SPFA();
	if(dis[n]>2*n)puts("-1");
	else printf("%d\n",dis[n]/2);
}