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<algorithm>
#include<vector>
#include<queue>
#define lol(i,n) for(int i=0;i<n;i++)
#define mod 1000000007
typedef long long ll;
using namespace std;
 
#include<map>
typedef pair<int,int> P;
 
vector<vector<P> >v;//辺リスト(v[i]には辺が詰まっている
map<int,int> u[100010];//頂点リスト(u[頂点][路線]==vでの番号)
 
priority_queue<P,vector<P>,greater<P> >Q;
int dis[500010];
 
int main(){
	vector<P> emptyvector;
	v.push_back(emptyvector);
	
	int n,m;
	cin>>n>>m;
	int vsize=1;
	for(int i=1;i<=n;i++){
		v.push_back(emptyvector);
		u[i][-1]=vsize;
		vsize++;
	}
	while(m--){
		int a,b,train;
		cin>>a>>b>>train;
		lol(p,2){
			if(u[a][train]==0){
				v.push_back(emptyvector);
				v[vsize].push_back(make_pair(a,0));
				v[a].push_back(make_pair(vsize,1));
				
				u[a][train]=vsize;
				a=vsize;
				vsize++;
			}
			else{
				a=u[a][train];
			}
			swap(a,b);
		}
		v[a].push_back(make_pair(b,0));
		v[b].push_back(make_pair(a,0));
	}
	
	//↓ Dijkstra ↓
	lol(i,vsize)dis[i]=mod;
	Q.push(make_pair(0,1));
	while(!Q.empty()){
		int b=Q.top().first,a=Q.top().second;
		Q.pop();
		if(dis[a]<=b)continue;
		dis[a]=b;
		for(int i=0;i<v[a].size();i++){
			int pnt=v[a][i].first,cost=v[a][i].second+b;
			if(dis[pnt]<=cost)continue;
			Q.push(make_pair(v[a][i].second+b,v[a][i].first));
		}
	}
	int tmp=dis[u[n][-1]];
	cout<<(tmp==mod?-1:tmp)<<endl;
	return 0;
}