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;
inline int read()
{
    int data=0;int w=1; char ch=0;
    ch=getchar();
    while(ch!='-' && (ch<'0' || ch>'9')) ch=getchar();
    if(ch=='-') w=-1,ch=getchar();
    while(ch>='0' && ch<='9') data=(data<<3)+(data<<1)+ch-'0',ch=getchar();
    return data*w;
}
const int N=1e5+10;
const int inf=1e9;
int n,m;
struct node{
	int v,w,nxt;
}e[N<<2];
int fir[N],cnt=0;
int dis[N],isin[N];
inline void add(int u,int v,int w){
	e[++cnt]=(node){v,w,fir[u]};fir[u]=cnt;
}
priority_queue<pair<int,int> >q;
set<int>c[N];
inline void dj(){
	for(int i=1;i<=n;i++) dis[i]=inf;
	memset(isin,0,sizeof(isin));
    dis[1]=0;
    q.push(make_pair(0,1));
    while(!q.empty()){
    	int u=q.top().second;q.pop();
    	if(isin[u]) continue;
    	isin[u]=1;
    	for(int i=fir[u];i;i=e[i].nxt){
    		int v=e[i].v,w=c[u].count(e[i].w)?0:1;
    		if(dis[v]>dis[u]+w){
    			dis[v]=dis[u]+w;
    			q.push(make_pair(-dis[v],v));
    			c[v].clear();
    			c[v].insert(e[i].w);
			}
			else if(dis[v]==dis[u]+w){
				c[v].insert(e[i].w);
			}
		}
	}
}
int main(){
	n=read();m=read();
	if(m==0) {
		puts("-1");return 0;
	}
	for(int i=1;i<=m;i++){
		int x=read(),y=read(),v=read();
		add(x,y,v);add(y,x,v);
	}
	dj();
	printf("%d\n",dis[n]==inf?-1:dis[n]);
	
	return 0;
}