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;

const int N=1001000;
int fa[N],n,m,q[N],d[N],a[N],bl[N];
vector<int>g[N];
vector<pair<int,int> >vec[N];
int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
void add(int x,int y){g[x].push_back(y);g[y].push_back(x);}

int bfs(int S,int T){
	for(int i=1;i<=n;++i)d[i]=-2;
	d[S]=0;int t=-1,w=0;q[0]=S;
	for(;t<w;){
		int x=q[++t];
		for(int i:g[x])if(d[i]==-2)d[i]=d[x]+1,q[++w]=i;
	}
	return d[T];
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n>>m;int S=1,T=n,mx=0;
	for(int i=1;i<=m;++i){
		int x,y,z;cin>>x>>y>>z;mx=max(mx,z);
		vec[z].emplace_back(x,y);
	}
	for(int i=1;i<=mx;++i)if(!vec[i].empty()){
		int cnt=0;
		for(auto j:vec[i])a[++cnt]=j.first,a[++cnt]=j.second;
		sort(a+1,a+cnt+1);cnt=unique(a+1,a+cnt+1)-a-1;
		for(int j=1;j<=cnt;++j)fa[a[j]]=a[j];
		for(auto j:vec[i])fa[find(j.first)]=find(j.second);
		for(int j=1;j<=cnt;++j)if(fa[a[j]]==a[j])bl[a[j]]=++n;
		for(int j=1;j<=cnt;++j)add(a[j],bl[find(a[j])]);
	}
	cout<<(bfs(S,T)/2)<<'\n';
	return 0;
}