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>
#define N 100010
#define M 400010
using namespace std;
template<typename T>inline void read(T &x)
{
	x=0;
	static int p;p=1;
	static char c;c=getchar();
	while(!isdigit(c)){if(c=='-')p=-1;c=getchar();}
	while(isdigit(c)) {x=(x<<1)+(x<<3)+(c-'0');c=getchar();}
	x*=p;
}
int n,m;
int e;
int to[M],beg[N],nex[M],Belong[M];
void add(int x,int y,int c)
{
	to[++e]=y;
	nex[e]=beg[x];
	beg[x]=e;
	Belong[e]=c;
}
struct node
{
	int dis,s;
	bool operator <(const node &ss)const { return dis==ss.dis?s<ss.s:dis<ss.dis; }
};
int dis[N],vis[N];
set<int>C[N*10];
void Dijkstra()
{
	priority_queue<node>q;
	memset(dis,0x3f,sizeof dis);
	dis[1]=0;
	q.push((node){0,1});
	while(!q.empty())
	{
		node k=q.top();q.pop();
		if(vis[k.s])continue;
		vis[k.s]=1;
		for(int i=beg[k.s];i;i=nex[i])
		{
			int y=to[i];
			int w=C[k.s].count(Belong[i])?0:1;
			if(dis[y]>dis[k.s]+w)
			{
				dis[y]=dis[k.s]+w;
				C[y].clear();C[y].insert(Belong[i]);
				q.push((node){-dis[y],y});
			}
			else if(dis[y]==dis[k.s]+w)C[y].insert(Belong[i]);
		}
	}
	printf("%d\n",dis[n]==0x3f3f3f3f?-1:dis[n]);
}
int main()
{
	read(n);read(m);
	for(int i=1;i<=m;i++)
	{
		static int x,y,c;
		read(x);read(y);read(c);
		add(x,y,c);add(y,x,c);
	}
	Dijkstra();
	return 0;
}