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<cctype>
#include<cstring>
#include<map>
#include<queue>
using namespace std;
void read(int &v)
{
	int f;char ch;
	while(!isdigit(ch=getchar())&&ch!='-'); ch=='-'?(f=-1,v=0):(f=1,v=ch-'0');
	while(isdigit(ch=getchar())) v=v*10+ch-'0';v*=f;
}
const int N=5e5+5,M=1e6+5;
struct sd{
	int next,to,w;
}edge[M<<1];
map<int,int> mp[N];
deque<int> que;
int head[N],dis[N],n,m,cnt,num;
int get(int x,int y)
{
	if(mp[x].count(y))
	return mp[x][y];
	return mp[x][y]=++cnt;
}
void add_edge(int from,int to,int w)
{
	edge[++num].next=head[from];
	edge[num].to=to;
	edge[num].w=w;
	head[from]=num;
}
int main()
{
	read(n),read(m);
	for(int i=1;i<=n;i++) get(i,0);
	for(int i=1;i<=m;i++)
	{
		int x,y,z;
		read(x),read(y),read(z);
		int x0=get(x,0),y0=get(y,0),xz=get(x,z),yz=get(y,z);
		add_edge(xz,yz,0),add_edge(yz,xz,0);
		add_edge(x0,xz,1),add_edge(xz,x0,1);
		add_edge(y0,yz,1),add_edge(yz,y0,1);
	}
	memset(dis,-1,sizeof(dis));dis[1]=0;
	que.push_back(1);
	while(!que.empty())
	{
		int v=que.front();que.pop_front();
		for(int i=head[v];i;i=edge[i].next)
		{
			int to=edge[i].to;
			if(dis[to]==-1||dis[to]>dis[v]+edge[i].w)
			{
				dis[to]=dis[v]+edge[i].w;
				if(edge[i].w==1) que.push_front(to);
				else que.push_back(to);
			}
		}
	}
	printf("%d",dis[n]>>1);
	return 0;
}