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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#include <tuple>
#define mkp make_pair
#define mkt make_tuple
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
typedef long long ll;
const ll MOD=1e9+7;

int N,M;
vector<pair<int,int>> g[200020];

const int ex=-1;
const int INF=1e9;

int dijkstra(int s,int t){
    vector<int> dist(2*N,0);
    rep(i,2*N) dist[i]=INF;
    dist[s]=0;
    priority_queue<tuple<int,int,int>,vector<tuple<int,int,int>>,greater<tuple<int,int,int>>> PQ;
    PQ.push(mkt(0,s,MOD));

    while(!PQ.empty()){
        int cost,now,pre;
        auto f=PQ.top();
        PQ.pop();

        cost=get<0>(f);
        now=get<1>(f);
        pre=get<2>(f);

        if(dist[now]<cost) continue;
        if(now==t) return dist[t];

        for(auto to:g[now]){
            int nex,c;
            tie(nex,c)=to;

            int neco=0;
            //if(pre==c||c==ex||pre==ex){
                if(c==ex) neco=1;
                else if(pre!=ex&&c!=pre) neco=1;
                if(dist[nex]>dist[now]+neco){
                    dist[nex]=dist[now]+neco;
                    PQ.push(mkt(dist[nex],nex,c));
                }
            //}
        }
    }

    return -1;
}

int main(){
  cin>>N>>M;
  for(int i=0;i<M;i++){
      int a,b,c;
      cin>>a>>b>>c;
      a--;b--;
      g[a].push_back(mkp(b,c));
      g[b].push_back(mkp(a,c));
  }

  for(int i=0;i<N;i++){
      for(int j=0;j<g[i].size();j++){
          int nex,c;
          tie(nex,c)=g[i][j];
          g[i+N].push_back(g[i][j]);
          g[i+N].push_back(mkp(nex+N,c));
      }
      g[i].push_back(mkp(i+N,ex));
  }

  int ans=dijkstra(0,N-1);
  cout<<ans<<endl;

  return 0;
}