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>
using namespace std;

typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;
typedef pair<LL, LL> PLL;

#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SORT(c) sort((c).begin(),(c).end())

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

#define FF first
#define SS second
template<class S, class T>
istream& operator>>(istream& is, pair<S,T>& p){
  return is >> p.FF >> p.SS;
}

const double EPS = 1e-10;
const double PI  = acos(-1.0);
const LL MOD = 1e9+7;
const int INF = 1e9;

int N, M;
struct Edge{
  int to, cost, type;

  Edge(int t, int c, int type): to(t), cost(c), type(type)
  {}
  bool operator>(const Edge& rhs) const{
	return cost > rhs.cost;
  }
  bool operator<(const Edge& rhs) const{
	return cost < rhs.cost;
  }

};
typedef vector< vector<Edge> > Graph;

// 無向グラフのとき
void add_edge(Graph& graph, int u, int v, int cost, int type){
  graph[u].push_back(Edge(v,cost,type));
  graph[v].push_back(Edge(u,cost,type));
}

int Dijkstra(const Graph& edges, map<PII,int>& d, int s){
  using Type = pair<int,PII>;
  priority_queue<Type, vector<Type>, greater<Type> > pq;
  d[MP(s,-1)] = 0;
  pq.push(MP(0,MP(s,-1)));
  int ans = INF;

  while(!pq.empty()){
	Type pii = pq.top(); pq.pop();
	PII p = pii.second;
	int v = p.first;
	if(d[p] < pii.first) continue;
	if(v == N-1) ans = min(ans, d[p]);
	for(unsigned int i=0;i<edges[v].size();++i){
	  const Edge& e = edges[v][i];
	  int nc = pii.first + (e.type != p.second? 1: 0);
	  PII next(e.to, e.type);
	  if(!d.count(next) || d[next] > nc){
		d[next] = nc;
		pq.push(MP(nc,next));
	  }
	}
  }

  return ans;
}


int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  cin >> N >> M;
  Graph G(N);
  REP(i,M){
	int a, b, c; cin >> a >> b >> c;
	--a; --b;
	add_edge(G, a, b, 0, c);
  }

  map<PII,int> memo;
  int ans = Dijkstra(G, memo, 0);
  cout << (ans==INF?-1:ans) << endl;;
  
  return 0;
}