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>
#define M 1000010
using namespace std;
int read()
{
	char ch=getchar();int x=0;
	while(ch>'9'||ch<'0') ch=getchar();
	while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
	return x;
}
struct point{
	int to,next,dis;
}e[M<<1];
int n,m,num,cnt;
int head[M],dis[M];bool vis[M];
void add(int from,int to,int dis)
{
	e[++num].next=head[from];
	e[num].to=to;
	e[num].dis=dis;
	head[from]=num;
}
void SPFA()
{
	memset(dis,127,sizeof(dis));
	queue<int>q;
	q.push(1);
	vis[1]=true;
	dis[1]=0;
	while(!q.empty())
	{
		int x=q.front(); q.pop();
		for(int i=head[x];i;i=e[i].next)
		{
			int to=e[i].to;
			if(dis[to]>dis[x]+e[i].dis)
			{
				dis[to]=dis[x]+e[i].dis;
				if(!vis[to])
				{
					vis[to]=true;
					q.push(to);
				}
			}
		}
		vis[x]=false;
	}
}
map<pair<int,int>,int>P;
int get(int x,int y)
{
	if(P.find(make_pair(x,y))!=P.end()) return P[make_pair(x,y)];
	else return P[make_pair(x,y)]=++cnt;
}
int main()
{
	n=cnt=read(); m=read();
	for(int i=1;i<=m;i++)
	{
		int a=read(),b=read(),c=read();
		int id1=get(a,c),id2=get(b,c);
		add(id1,id2,0); add(id2,id1,0);
		add(a,id1,1); add(id1,a,1);
		add(b,id2,1); add(id2,b,1);
	}
	SPFA();
	printf("%d\n",dis[n]>2*n?-1:dis[n]/2);
	return 0;
}