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:
#pragma GCC optimize("-Ofast","-funroll-all-loops")
#include<bits/stdc++.h>
//#define int long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=1e5+10,M=1e6+10;
int n,m,d[N],vis[N];
int head[N],nex[M],to[M],w[M],tot;
priority_queue<pair<int,int>> q;	set<int> s[N];
inline void ade(int a,int b,int c){
	to[++tot]=b; nex[tot]=head[a]; w[tot]=c; head[a]=tot;
}
inline void add(int a,int b,int c){ade(a,b,c);	ade(b,a,c);}
void Dijkstra(){
	memset(d,inf,sizeof d);	d[1]=0; q.push({0,1});
	while(q.size()){
		int u=q.top().second;	q.pop();
		if(vis[u])	continue;	vis[u]=1;
		for(int i=head[u],val;i;i=nex[i]){
			val=(s[u].find(w[i])!=s[u].end()?0:1);
			if(d[to[i]]>d[u]+val){
				d[to[i]]=d[u]+val;	q.push({-d[to[i]],to[i]});
				s[to[i]].clear();	s[to[i]].insert(w[i]);
			}else if(d[to[i]]==d[u]+val)	s[to[i]].insert(w[i]);
		}
	}
}
signed main(){
	cin>>n>>m;
	for(int i=1,a,b,c;i<=m;i++)	scanf("%d %d %d",&a,&b,&c),add(a,b,c);
	Dijkstra();
	if(d[n]==inf)	d[n]=-1;
	cout<<d[n]<<endl;
	return 0;
}