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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000000000002

int main(){
	
	int N,M;
	cin>>N>>M;
	
	vector<map<int,vector<int>>> E(N);
	
	for(int i=0;i<M;i++){
		int p,q,c;
		scanf("%d %d %d",&p,&q,&c);
		p--;q--;
		E[p][c].push_back(q);
		E[q][c].push_back(p);
	}
	
	map<pair<int,int>,int> dis;
	dis[make_pair(0,0)]=0;
	deque<pair<int,int>> Q;
	Q.push_back(make_pair(0,0));
	set<pair<int,int>> visited;
	while(Q.size()!=0){
		int x = Q.back().first;
		int u = Q.back().second;
		Q.pop_back();
		if(visited.count(make_pair(x,u)))continue;
		if(x==0){
			for(auto a:E[u]){
				int v = a.first;
				if(dis.count(make_pair(v,u)))continue;
				dis[make_pair(v,u)] = dis[make_pair(x,u)]+1;
				Q.emplace_front(v,u);
			}
		}
		else{
			for(int i=0;i<E[u][x].size();i++){
				int v = E[u][x][i];
				if(dis.count(make_pair(x,v))&&dis[make_pair(x,v)]<=dis[make_pair(x,u)])continue;
				dis[make_pair(x,v)]=dis[make_pair(x,u)];
				Q.emplace_back(x,v);
			}
			int v = u;
			if(dis.count(make_pair(0,v))&&dis[make_pair(0,v)]<=dis[make_pair(x,u)])continue;
			dis[make_pair(0,v)]=dis[make_pair(x,u)];
			Q.emplace_back(0,v);
		}
	}
	
	pair<int,int> ans(0,N-1);
	if(dis.count(ans))cout<<dis[ans]<<endl;
	else cout<<-1<<endl;
	
	return 0;
}