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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;

template<class T> struct edge{
	int to;
	T wt;
	edge(int to,const T& wt):to(to),wt(wt){}
};
template<class T> using weighted_graph=vector<vector<edge<T>>>;

template<class T>
vector<T> Dijkstra(const weighted_graph<T>& G,int s){
	const T INF=numeric_limits<T>::max();
	int n=G.size();
	vector<T> d(n,INF); d[s]=0;
	priority_queue<pair<T,int>> Q; Q.emplace(0,s);
	while(!Q.empty()){
		T d0=-Q.top().first;
		int u=Q.top().second; Q.pop();
		if(d0>d[u]) continue;
		for(const auto& e:G[u]){
			int v=e.to;
			if(d[v]>d[u]+e.wt) d[v]=d[u]+e.wt, Q.emplace(-d[v],v);
		}
	}
	return d;
}

int main(){
	int n,m; scanf("%d%d",&n,&m);

	int k=n;
	vector<map<int,int>> f(n); // f[u][color] := (vertex id)

	vector<tuple<int,int,int>> E(m);
	rep(i,m){
		int u,v,c; scanf("%d%d%d",&u,&v,&c); u--; v--; c--;
		E[i]=make_tuple(u,v,c);
		if(f[u].count(c)==0) f[u][c]=k++;
		if(f[v].count(c)==0) f[v][c]=k++;
	}

	weighted_graph<int> G(k);
	rep(u,n) for(auto p:f[u]) {
		int i=p.second;
		G[u].emplace_back(i,1);
		G[i].emplace_back(u,1);
	}
	for(auto e:E){
		int u,v,c; tie(u,v,c)=e;
		int i=f[u][c],j=f[v][c];
		G[i].emplace_back(j,0);
		G[j].emplace_back(i,0);
	}

	int res=Dijkstra(G,0)[n-1];
	printf("%d\n",res<1e8?res/2:-1);

	return 0;
}