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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define int long long
using namespace std;

typedef pair<int,int> P;
typedef pair<int,P> P1;

const int INF = 1e9;

int n, m;
vector<P> G[100005];

unordered_map<int,int> d[100005];
int mind[100005];

int dijkstra(){
  
  for(int i=0;i<n;i++) mind[i] = INF;
  
  priority_queue<P1,vector<P1>,greater<P1> > q;

  mind[0] = 0;
  d[0][0] = 0;
  q.push(P1( 0, P( 0, 0 ) ));
  
  while(!q.empty()){
    
    P1 tmp = q.top(); q.pop();
    
    int cost = tmp.first;
    int c = tmp.second.first;
    int node = tmp.second.second;
    
    if( node == n-1 ) return cost;
    
    if( d[node].count(c) && d[node][c] < cost ) continue;
    
    for(P p : G[node] ){
      
      int nnode = p.first;
      int nc = p.second;
      int ncost = cost;
      
      if( c != nc ) ncost++;
      
      if( mind[nnode] + 2 <= ncost ) continue;
      
      if( d[nnode].count(nc) == 0 ){
	mind[nnode] = min( mind[nnode], ncost );
	d[nnode][nc] = ncost;
	q.push(P1( ncost, P( nc, nnode ) ));
      }
      else if( d[nnode][nc] > ncost ){
	mind[nnode] = min( mind[nnode], ncost );
	d[nnode][nc] = ncost;
	q.push(P1( ncost, P( nc, nnode ) ));
      }
      
    }
    
  }
  
}

int used[100005];

bool check(int x){
  
  if( x == n-1 ) return true;
  
  if( used[x] ) return false;
  used[x] = 1;
  
  bool res = false;

  for(P p : G[x] ) res |= check(p.first);

  return res;
}

signed main(){
  
  cin>>n>>m;
  
  for(int i=0;i<m;i++){
    int a, b, c;
    cin>>a>>b>>c;
    G[a-1].push_back(P(b-1,c));
    G[b-1].push_back(P(a-1,c));
  }

  if( !check(0) ) cout<<-1<<endl;
  else cout<<dijkstra()<<endl;
  
  return 0;
}