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 <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <map>
#include<set>
#define fre(x) freopen(x".in","r",stdin),freopen(x".out","w",stdout)
using namespace std;

inline int read(){
	int sum=0,f=1;char ch=getchar();
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){sum=sum*10+ch-'0';ch=getchar();}
	return f*sum;
}

const int N=2000005;
struct edge{int to,next,c;}e[N];
int n,m,d[N],v[N];
int head[N],cnt=0;
//map < pair<int,int>,int >mp;
set<int>se[N];
queue <int> que;

inline void add(int x,int y,int w){
	e[++cnt]=(edge){head[x],y,w};head[x]=cnt;
}

signed main(){
	n=read();m=read();
	for(int i=1;i<=m;i++){
		int x=read(),y=read(),w=read();
		add(x,y,w);add(y,x,w);
	}
	memset(d,0x3f,sizeof(d));d[1]=0;
	que.push(1);
	while(!que.empty()){
		int x=que.front();que.pop();
//		cout<<"x="<<x<<endl;
		v[x]=0;
		for(int i=head[x];i;i=e[i].to){
			int k=e[i].next;
			int w=se[x].count(e[i].c)^1;
			if(d[k]>d[x]+w){
				d[k]=d[x]+w;se[k].clear();
// 				mp[make_pair(k,e[i].c)]=1;
				se[k].insert(e[i].c);
				if(!v[k]) que.push(k),v[k]=1;
			}
			if(d[k]==d[x]+w){
				if(se[k].find(e[i].c)==se[k].end()){
					if(!v[k]) que.push(k),v[k]=1;
				}
				se[k].insert(e[i].c);
			}
		}	
	}
	if(d[n]==1061109567) puts("-1");
	else printf("%d\n",d[n]);
	return 0;
}