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 int long long
typedef pair<int,int> P;
#define MAX 114514
vector<P> G[MAX*5];
map<int,int> M[MAX];
int dist[MAX*5];
signed main(){
  int n,m;
  cin>>n>>m;
  int cnt=n;
  for(int i=0;i<m;i++){
    int p,q,c;
    cin>>p>>q>>c;
    p--;q--;
    auto add_edge = [&](int x){
      if(!M[x].count(c)){
	M[x][c]=cnt++;
	G[M[x][c]].push_back(P(x,0));
	G[x].push_back(P(M[x][c],1));
      }
    };
    add_edge(p);
    add_edge(q);
    G[M[p][c]].push_back(P(M[q][c],0));
    G[M[q][c]].push_back(P(M[p][c],0));
  }
  memset(dist,-1,sizeof(dist));
  priority_queue<P,vector<P>,greater<P> > q;
  dist[0]=0;
  q.push(P(dist[0],0));
  while(!q.empty()){
    P p=q.top();q.pop();
    int d=p.first,v=p.second;
    if(dist[v]<d) continue;
    for(P e:G[v]){
      int u=e.first,c=e.second;
      if(dist[u]<0||dist[u]>d+c){
	dist[u]=d+c;
	q.push(P(dist[u],u));
      }
    }
  }
  cout<<dist[n-1]<<endl;
  return 0;
}