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++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>

const double pi = 3.141592653589793238462643383279;
using namespace std;
using namespace __gnu_pbds;
//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<long long, long long> PLL;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
 
 
//container util
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
 
 
//repetition
//------------------------------------------
#define FOR(i,s,n) for(int i=s;i<(int)n;++i)
#define REP(i,n) FOR(i,0,n)
#define MOD 1000000007
 
 
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
 
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
 
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)

const ll INF = -(1LL<<60);

const long double EPS = 1e-6, PI = acos((long double)-1);

//ここから編集

vector<vector<pair<int, int>>> g(101010);
map<int, int> dist[101010];
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(7);

  int N, M; cin >> N >> M;

  REP(i,M){
    int p, q, c;
    cin >> p >> q >> c;
    p--;
    q--;
    g[p].push_back({q, c});
    g[q].push_back({p, c});
  }

  priority_queue<pair<int, pair<int, int>>, vector<pair<int, pair<int, int>>>, greater<pair<int, pair<int, int>>>> q;
  q.push(make_pair(0, make_pair(0, -1)));

  dist[0][-1] = 0;

  while(q.size()){

    int v, prev;
    auto p = q.top();
    q.pop();
    int cost;
    v = p.second.first;
    cost = p.first;
    prev = p.second.second;
    if(dist[v][prev] < cost) continue;

    for(auto e: g[v]){
      int u = e.first;
      int c = e.second;

      if(prev != c){
        if(dist[u].find(c) == dist[u].end()){
          dist[u][c] = dist[v][prev] + 1;
          q.push(make_pair(dist[u][c], make_pair(u, c)));
        }else if(dist[u][c] > dist[v][prev] + 1){
          dist[u][c] = dist[v][prev] + 1;
          q.push(make_pair(dist[u][c], make_pair(u, c)));
        }
      }else{
        if(dist[u].find(c) == dist[u].end()){
          dist[u][c] = dist[v][prev];
          q.push(make_pair(dist[u][c], make_pair(u, c)));
        }else if(dist[u][c] > dist[v][prev]){
          dist[u][c] = dist[v][prev];
          q.push(make_pair(dist[u][c], make_pair(u, c)));
        }
      }
    }
  }

  int ans = INT_MAX;
  for(auto e: dist[N-1]){
    ans = min(ans, e.second);
  }
  if(ans == INT_MAX) ans = -1;
  cout << ans << endl;
  return 0;
}