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;
typedef pair<int,int> P;

#define MAX_N 100005
int N,M;
map<  int, map<int , vector<int> >  > G;
map<int, map<int,bool> > visited;
int d[MAX_N];

vector<int> v;

void dfs(int pos,int ID){
  if(visited[pos][ID])return;
  visited[pos][ID]=true;
  v.push_back(pos);
  for(int i=0;i<(int)G[pos][ID].size();i++){
    dfs(G[pos][ID][i],ID);
  }
}

int main(){
  scanf("%d %d",&N,&M);
  for(int i=0;i<=N;i++)d[i]=1e8;
  
  for(int i=0;i<M;i++){
    int a,b,c;
    scanf("%d %d %d",&a,&b,&c);
    G[a][c].push_back( b );
    G[b][c].push_back( a );
  }


  priority_queue< P , vector<P> , greater<P> > Q;
  Q.push(P(0,1));
  d[1]=0;
  while(!Q.empty()){
    P p=Q.top();Q.pop();
    int pos=p.second;
    int cost=p.first;
    if(d[pos]<cost)continue;
    map<int, vector<int> > &g=G[pos];
    for( auto &e : g ){
      if( visited[pos][e.first] )continue;
      v.clear();
      dfs(pos,e.first);
      for(int i=0;i<(int)v.size();i++){
        int to=v[i];
        if( d[to] > cost+1 ){
          d[to]=cost+1;
          Q.push(P(cost+1,to));
        }
      }
    }
  }

  if(d[N]==1e8)cout<<-1<<endl;
  else cout<<d[N]<<endl;
  
  return 0;
}