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;

//repetition
#define FOR(i,a,b) for(ll i=(a);i<(b);++i)
#define rep(i, n) for(ll i = 0; i < (ll)(n); i++)

//container util
#define all(x) (x).begin(),(x).end()

//typedef
typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<ll> VLL;
typedef vector<VLL> VVLL;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<ll, ll> PLL;

//const value
//const ll MOD = 1e9 + 7;
//const int dx[] = {0,1,0,-1};//{0,0,1,1,1,-1,-1,-1};
//const int dy[] = {1,0,-1,0};//{1,-1,0,1,-1,0,1,-1};

//conversion
inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
inline ll toLL(string s) {ll v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

int n,m;
const int INF = 1e9;
typedef tuple<int,int> State;

vector<PII> edge[3010101];
map<PII, int> sid;
int lastId = 0;

// 頂点と色を引数に持って新しく追加する頂点との間にエッジを張る
// 今まで出てきた端点と同じものがあったら同じグループに追加する
int getid(int v, int c){
  PII p(v,c);
  if( sid.find(p) !=  sid.end()) return sid[p];
  return sid[p] = lastId++;
}

void addEdge(int a, int b, int c){
  // 同じ辺の要素はコスト0の辺を貼ったのでそれぞれの端点で独立に考える
  int v = getid(a,c), u = getid(b,c);
  edge[v].emplace_back(u,0);
  edge[u].emplace_back(v,0);
}

void dijkstra(){
  int dist[n + lastId];
  rep(i,n+lastId) dist[i] = INF;
  dist[0] = 0;
  priority_queue< State, vector<State>, greater<State>> que;
  que.push(State(0,0));
  while(que.empty() == false){
    State p = que.top(); que.pop();
    int cdist = get<0>(p), current = get<1>(p);

    if(dist[current] < cdist) continue;

    for(PII e: edge[current]){
      if(dist[e.first] > cdist + e.second){
          que.push(State(cdist + e.second,e.first));
          dist[e.first] = e.second + cdist;
      }
    }
  }

  if(dist[n-1] == INF){
    cout << -1 << endl;
    return;
  }else cout << dist[n-1]/2 << endl;

}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;

  int a[m],b[m],c[m];

  rep(i,m){
    cin >> a[i] >> b[i] >> c[i];
    a[i]--,b[i]--;
    addEdge(a[i],b[i],c[i]);
  }

  // 作成した頂点と入力の辺の端点からコスト1の辺を張る
  rep(i,m){
    edge[getid(a[i], c[i])].emplace_back(lastId+a[i], 1);
    edge[getid(b[i], c[i])].emplace_back(lastId+b[i], 1);
    edge[lastId+a[i]].emplace_back(getid(a[i], c[i]), 1);
    edge[lastId+b[i]].emplace_back(getid(b[i], c[i]), 1);
  }

  dijkstra();
  return 0;
}