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;

typedef long long lli;
typedef vector<lli> vll;
typedef vector<bool> vbl;
typedef vector<vector<lli>> mat;
typedef unordered_map<lli,multimap<lli,lli>> graph;

lli n,m;
graph g;
graph h;
priority_queue<vll,vector<vll>,greater<vll>> q;
set<vll> used;

int main(){
  cin >> n >> m;
  for(lli i = 0;i < m;i++){
    lli p,q,c;
    cin >> p >> q >> c;
    g[p].insert(make_pair(q,c));
    g[p].insert(make_pair(p,c));
    h[c].insert(make_pair(p,q));
    h[c].insert(make_pair(q,p));
  }
  q.push(vll({0,1,0}));
  while(!q.empty()){
    lli cost = q.top()[0];
    lli station = q.top()[1];
    lli company = q.top()[2];
    q.pop();
    if(station == n){
      cout << cost << endl;
      return 0;
    }
    if(used.count(vll({station,company})) != 0) continue;
    used.insert(vll({station,company}));

    if(company == 0){
      for(auto itr : g[station]){
        if(used.count(vll({itr.first,itr.second})) == 0) q.push(vll({cost+1,itr.first,itr.second}));
      }
    }else{
      if(used.count(vll({station,0})) == 0) q.push(vll({cost,station,0}));
      for(auto itr = h[company].lower_bound(station);itr != h[company].upper_bound(station);itr++){
        if(used.count(vll({itr->second,company})) == 0) q.push(vll({cost,itr->second,company}));
      }      
    }
  }
  cout << -1 << endl;
  return 0;
}