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 <algorithm>
#include <queue>
#include <vector>
using namespace std;
struct Node
{
	int x,y;
};
struct Edge
{
	int to,next;
}e[5000005];
vector<Node>a[1000005];
int f[100005],rt[100005],d[400005],h[400005],cnt;
inline void Addedge(int x,int y)
{
	e[++cnt]=(Edge){y,h[x]};h[x]=cnt;return;
}
inline int GetFather(int x)
{
	if(f[x]==x)return x;
	return f[x]=GetFather(f[x]);
}
inline void Merge(int x,int y)
{
	x=GetFather(x);
	y=GetFather(y);
	if(x!=y)f[x]=y;
	return;
}
queue<int>q;
int main(void)
{
	int i,j,x,y,n,m,c,id;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)
	{
		scanf("%d%d%d",&x,&y,&c);
		a[c].push_back((Node){x,y});
	}
	id=n;
	for(i=1;i<=1000000;++i)
	{
		for(j=0;j<a[i].size();++j)
		{
			x=a[i][j].x;y=a[i][j].y;
			f[x]=x;f[y]=y;
			rt[x]=rt[y]=0;
		}
		for(j=0;j<a[i].size();++j)Merge(a[i][j].x,a[i][j].y);
		for(j=0;j<a[i].size();++j)
		{
			x=a[i][j].x;y=a[i][j].y;
			if(!rt[GetFather(x)])rt[GetFather(x)]=++id;
			Addedge(x,rt[GetFather(x)]);
			Addedge(rt[GetFather(x)],x);
			Addedge(y,rt[GetFather(y)]);
			Addedge(rt[GetFather(y)],y);
		}
	}
	q.push(1);
	while(!q.empty())
	{
		x=q.front();q.pop();
		for(i=h[x];i;i=e[i].next)
		{
			y=e[i].to;
			if(!d[y]){d[y]=d[x]+1;q.push(y);}
		}
	}
	if(d[n])printf("%d\n",d[n]/2);
	else printf("-1\n");
	return 0;
}