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>
using namespace std;

int n,m;

const int maxn=1e5+5;

unordered_map<int,vector<int> >e[maxn];
set<int>g[maxn];//颜色 
unordered_map<int,int>d[maxn];

struct st
{
	int p,b,c;//点 颜色 花费 
	st(int x,int y,int z)
	{
		p=x;b=y;c=z;
	}
};
bool operator < (const st &a,const st &b)
{
	return a.c>b.c;
}
priority_queue<st>q;

int main()
{
	ios::sync_with_stdio(0);cin.tie(0);
	
	cin>>n>>m;
	for(int i=1;i<=m;i++)
	{
		int u,v,c;cin>>u>>v>>c;
		e[u][c].push_back(v);
		e[v][c].push_back(u);
		g[u].insert(c);
		g[v].insert(c);
	}
	q.push(st(1,0,0));d[1][0]=0;
	while(!q.empty())
	{
		st tmp=q.top();q.pop();
		if(d[tmp.p][tmp.b]!=tmp.c)continue;
		if(!d[tmp.p].count(0)||d[tmp.p][0]>tmp.c)
		{
			d[tmp.p][0]=tmp.c;
			q.push(st(tmp.p,0,tmp.c));
		}
		if(tmp.b==0)
		{
			for(auto v:g[tmp.p])
			{
				if(!d[tmp.p].count(v)||d[tmp.p][v]>tmp.c+1)
				{
					d[tmp.p][v]=tmp.c+1;
					q.push(st(tmp.p,v,tmp.c+1)); 
				}
			}
		}
		else
		{
			for(auto v:e[tmp.p][tmp.b])
			{
				if(!d[v].count(tmp.b)||d[v][tmp.b]>tmp.c)
				{
					d[v][tmp.b]=tmp.c;
					q.push(st(v,tmp.b,tmp.c));
				}
			}
		}
	}
	int ans=INT_MAX;
	for(auto i:d[n])
	{
		ans=min(ans,i.second);
	}
	cout<<(ans==INT_MAX?-1:ans)<<'\n'; 
	
	return 0;
}