Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define rep(i,b) FOR(i,0,b)
#define INF 1e9
#define dump(x) cerr<<#x<<"="<<x<<endl
#define all(a) (a).begin(),(a).end()
typedef vector<int> V;
typedef vector<V> VV;
typedef vector<VV> VVV;
template <class T> void chmin(T & a, T const & b) { if (b < a) a = b; }
typedef pair<int,int> P;
using ll = long long;
const ll mod = LLONG_MAX;

struct data_t{
  //今の頂点
  int now;
  //コスト
  int cost;
  //前の会社
  int company;
  data_t(int now,int cost,int company):now(now),cost(cost),company(company){};

  bool operator > (const data_t &s)const{
    return (cost<s.cost);
  }
};

struct data_s{
  int to;
  int company;
  data_s(int to,int company):to(to),company(company){};
};




vector<data_s> edge[112345];
int N,M;
int d[112345];
void dijkstra(int s){
  priority_queue<data_t,vector<data_t>,greater<data_t>> que;
  fill(d,d+N,INF);
  d[s]=0;
  //コスト,今の点,経由してきた会社スターとは-1
  que.push(data_t(s,1,-1));

  while(!que.empty()){
   data_t p=que.top();que.pop();
   //今の頂点
   int v=p.now;
   for(int i=0;i<edge[v].size();i++){
     int next=edge[v][i].to;

     if(d[next]<=p.cost)continue;

     int nextCompany=edge[v][i].company;
     //-1はワイルドカード
     if(p.company==nextCompany || p.company==-1){
       dump(d[next]);
       d[next]=p.cost;
     }else{
       dump(d[next]);
       d[next]=p.cost+1;
     }
     que.push(data_t(next,d[next],nextCompany));

   }

  }
}
int main(){


 cin>>N>>M;
 rep(i,M){
   int a,b,c;

   cin>>a>>b>>c;
   a--;b--;
   edge[a].push_back(data_s(b,c));
   edge[b].push_back(data_s(a,c));


 }
 int s=0;
 dijkstra(s);
 if(d[N-1==INF])cout<<-1<<endl;
 else cout<<d[N-1]<<endl;

}