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:
/*********************
*  Author: xuziyuan  * 
*********************/

#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=1e9+7;

int n,m,x,y,z,len,dist[500010];
vector <pii> g[500010];
map <pii,int> mp;
priority_queue <pii,vector <pii>,greater <pii> > q;

void addedge(int pos,int pos2,int val,int val2)
{
	g[pos].pb(mpr(pos2,val));
	g[pos2].pb(mpr(pos,val2));
}

int getdownid(int pos,int val)
{
	if(mp.find(mpr(pos,val))!=mp.end()) return mp[mpr(pos,val)];
	int downid=++len;
	addedge(pos,downid,1,0);
	return mp[mpr(pos,val)]=downid;
}

int main()
{
	cin>>n>>m;
	len=n;
	rep(i,m)
	{
		scanf("%d%d%d",&x,&y,&z);
		int pos=getdownid(x,z),pos2=getdownid(y,z);
		addedge(pos,pos2,0,0);
	}
	q.push(mpr(0,1));
	rep(i,500005) dist[i]=1e9;
	dist[1]=0;
	while(!q.empty())
	{
		pii t=q.top();
		q.pop();
		rep(i,g[t.se].size())
		{
			if(dist[g[t.se][i].fi]<=t.fi+g[t.se][i].se) continue;
			dist[g[t.se][i].fi]=t.fi+g[t.se][i].se;
			q.push(mpr(dist[g[t.se][i].fi],g[t.se][i].fi));
		}
	}
	if(dist[n]==1e9) puts("-1");
	else cout<<dist[n]<<endl;
	return 0;
}