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>
#include<queue>
using namespace std;
const int N=1000010;
vector<int>e[N],f[N];
int n,m;
struct edge
{
	int x,y,c;
	bool operator < (const edge a)const
	{
		return c<a.c;
	}
}q[N];
int col[N],now[N],cnt,fi=N,la=N;
int qu[N<<1];
int dis[N];
void newnode(int x,int c)
{
	now[x]=++cnt; col[x]=c;
	e[x].push_back(cnt); f[x].push_back(1);
	e[cnt].push_back(x); f[cnt].push_back(0);
}
void bfs()
{
	memset(dis,0,sizeof(dis));
	qu[fi]=1; 
	while(fi<=la)
	{
		int x=qu[fi++];//printf("%d\n",x);
		for(int i=0;i<e[x].size();i++)
			if((!dis[e[x][i]] || dis[e[x][i]]>dis[x]+f[x][i]) && e[x][i]!=1)
			{
				dis[e[x][i]]=dis[x]+f[x][i];
				if(f[x][i]==0)
					qu[--fi]=e[x][i];
				else
					qu[++la]=e[x][i];
			}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
		scanf("%d%d%d",&q[i].x,&q[i].y,&q[i].c);
	sort(q+1,q+m+1);
	cnt=n;
	for(int i=1;i<=m;i++)
	{
		int x=q[i].x,y=q[i].y,c=q[i].c;
		if(col[x]!=c)
			newnode(x,c);
		if(col[y]!=c)
			newnode(y,c);
		e[now[x]].push_back(now[y]);
		f[now[x]].push_back(0);
		e[now[y]].push_back(now[x]);
		f[now[y]].push_back(0);
	}
	bfs();
	if(dis[n])
		printf("%d\n",dis[n]);
	else
		printf("-1\n");
	return 0;
}