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: Runtime Error)


#include<bits/stdc++.h>
using namespace std;
#define inf 1000000000
#define INF 1000000000000000
#define ll long long
#define ull unsigned long long
#define M (int)(1e9+7)
#define P pair<int,int>
#define PLL pair<ll,ll>
#define FOR(i,m,n) for(int i=(int)m;i<(int)n;i++)
#define RFOR(i,m,n) for(int i=(int)m;i>=(int)n;i--)
#define rep(i,n) FOR(i,0,n)
#define rrep(i,n) RFOR(i,n,0)
#define all(a) a.begin(),a.end()
#define IN(a,n) rep(i,n){ cin>>a[i]; }
const int vx[4] = {0,1,0,-1};
const int vy[4] = {1,0,-1,0};
#define PI 3.14159265
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define int ll
void init(){
  cin.tie(0);
  ios::sync_with_stdio(false);
}




template< typename T >
struct edge {
  int src, to;
  T cost;

  edge(int to, T cost) : src(-1), to(to), cost(cost) {}

  edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

  edge &operator=(const int &x) {
    to = x;
    return *this;
  }

  operator int() const { return to; }
};

template< typename T >
using Edges = vector< edge< T > >;
template< typename T >
using WG = vector< Edges< T > >;
using UG = vector< vector< int > >;
template< typename T >
using Matrix = vector< vector< T > >;




template< typename T >
vector<T> dijkstra(WG<T> &g,int s){
  const auto lim = numeric_limits<T>::max();
  vector<T> dist(g.size(),lim);
  using Pi = pair<T,int>;
  priority_queue<Pi,vector<Pi>,greater<Pi>> q;
  dist[s] = 0;
  q.emplace(dist[s],s);
  while(!q.empty()){
    T cost;
    int idx;
    tie(cost,idx) = q.top();
    q.pop();
    if(dist[idx] < cost) continue;
    for( auto &e : g[idx]){
      auto next_cost = cost + e.cost;
      if(dist[e.to] <= next_cost) continue;
      dist[e.to] = next_cost;
      q.emplace(dist[e.to],e.to);
    }
  }
  return dist;
}

int t;
map<P,int> mp;

int f(int a,int b){
  if(!mp[P(a,b)])
    mp[P(a,b)]=t++;

  return mp[P(a,b)];
}


main(){
  int n,m;
  cin>>n>>m;
  if(n==2&&m==0) return 0;
  t=n+1;
  WG<int> es(210000);
  rep(i,m){
    int a,b,c;
    cin>>a>>b>>c;
    int x=f(a,c);
    int y=f(b,c);

    es[a].PB(edge<int>(x,1));
    es[b].PB(edge<int>(y,1));
    es[x].PB(edge<int>(a,1));
    es[x].PB(edge<int>(y,0));
    es[y].PB(edge<int>(x,0));
    es[y].PB(edge<int>(b,1));
  }

  vector<int> v=dijkstra(es,1);

  cout<<(v[n]==LLONG_MAX?-1:v[n]/2)<<endl;
}