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 <string>
#include <vector>
#include <set>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <queue>
#include <map>
#include <numeric>
#include <unordered_map>
#include <iomanip>
#include <functional>
#include <bitset>
#include <complex>

#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define all(x) (x).begin(),(x).end()

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

typedef long long ll;

using namespace std;
struct ed{int a,b,c;};
using P=pair<int,int>;

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n,m;
  cin>>n>>m;
  vector<unordered_map<int, int>>num(n);
  vector<ed>es(m);
  if(m==0){cout<<-1<<endl;return 0;}

  rep(i,m)
  {
    int a,b,c;
    cin>>a>>b>>c;
    a--;b--;c--;
    es[i]={a,b,c};
    num[a][c]++;
    num[b][c]++;
  }
  int cnt=0;
  rep(i,n)
  {
    int tn=cnt++;
    for(auto p:num[i])
    {
      num[i][p.first]=cnt++;
    }
    num[i][-1]=tn;
  }
  vector<int>dist(cnt+2,-1);
  vector<vector<P>>edge(cnt+2);
  rep(i,n)for(auto p:num[i])
  {
    if(p.first==-1)continue;
    int t=num[i][-1];
    int tn=p.second;
    edge[t].push_back({1,tn});
    edge[tn].push_back({0,t});
  }
  rep(i,m)
  {
    int to=num[es[i].a][es[i].c],from=num[es[i].b][es[i].c];
    edge[to].push_back({0,from});
    edge[from].push_back({0,to});
  }
  int start=num[0][-1],end=num[n-1][-1];
  priority_queue<P,vector<P>,greater<P>>q;
  q.push({0,start});
  while(!q.empty())
  {
    P tp=q.top();q.pop();
    if(dist[tp.second]!=-1)continue;
    if(tp.second==end){cout<<tp.first<<endl;return 0;}
    dist[tp.second]=tp.first;
    for(P nn:edge[tp.second])
    {
      if(dist[nn.second]!=-1)continue;
      q.push({nn.first+tp.first,nn.second});
    }
  }
  cout<<-1<<endl;
  //rep(i,dist.size())cout<<i<<" "<<dist[i]<<endl;
  //cout<<end<<endl;

  rep(i,edge.size())rep(j,edge[i].size())
  {
    cout<<i<<" "<<edge[i][j].first<<" "<<edge[i][j].second<<endl;
  }



  return 0;

}