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 INF = 1e9;

vector<pair<int,int> > g[100001],edge[500001];
map<int,int> M[100001];
bool visit[500001];
int stp;
int dist[500001];

void ADD(int a,int b,int x){
	edge[a].emplace_back(make_pair(b,x));
	edge[b].emplace_back(make_pair(a,x));
}

void addedge(int a,int b,int c,int d){
	ADD(a,b,1);ADD(b,c,0);ADD(c,d,1);
}

void dfs(int x){
	visit[x]=true;
	for(auto it:g[x]){
		int y=it.first,c=it.second;
		if(!M[x][c])
			M[x][c]=++stp;
		if(!M[y][c])
			M[y][c]=++stp;
		addedge(x,M[x][c],M[y][c],y);
		if(!visit[y])
			dfs(y);
	}
}

void Dijkstra(){
	memset(visit,0,sizeof visit);
	for(int i=1;i<=stp;i++)
		dist[i]=INF;
	priority_queue<pair<int,int>,vector<pair<int,int> >, greater<pair<int,int> > > pq;
	dist[1]=0;
	pq.push(make_pair(0,1));
	while(!pq.empty()){
		int from=pq.top().second;pq.pop();
		if(visit[from])	continue;
		visit[from]=true;
		for(auto it:edge[from]){
			int to=it.first,cost=it.second;
			if(dist[from]+cost<dist[to]){
				dist[to]=dist[from]+cost;
				pq.push(make_pair(dist[to],to));
			}
		}
	}

}

int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>n>>m;
	for(int i=1,x,y,c;i<=m;i++){
		cin>>x>>y>>c;
		g[x].emplace_back(make_pair(y,c));
		g[y].emplace_back(make_pair(x,c));
	}
	stp=n;
	dfs(1);
	Dijkstra();
	cout<<(dist[n]==INF?-1:dist[n]/2)<<endl;


}