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 int64;

struct UnionFind
{
  vector< int > data;

  UnionFind(int sz)
  {
    data.assign(sz, -1);
  }

  void unite(int x, int y)
  {
    x = find(x), y = find(y);
    if(x != y) {
      if(data[x] > data[y]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
  }

  int find(int k)
  {
    if(data[k] < 0) return (k);
    return (data[k] = find(data[k]));
  }

  int size(int k)
  {
    return (-data[find(k)]);
  }
};

queue < int > que[200001];
map< int, bool > v[100000];
bool visited[100000];
int N, M;
map< int, map< int, bool > > g[1000001];
vector< pair< int, int > > graph[100000];

void dfs(queue< int >& q, int idx, int cc)
{
  for(auto to : g[cc][idx]) {
    if(v[idx][to.first]++) continue;
    if(!visited[to.first]++) q.push(to.first);
    dfs(q, to.first, cc);
  }
}

int main()
{
  cin >> N >> M;
  for(int i = 0; i < M; i++) {
    int p, q, c;
    cin >> p >> q >> c;
    --p, --q;
    g[c][p][q]++;
    g[c][q][p]++;
    graph[p].push_back({q, c});
    graph[q].push_back({p, c});
  }

  que[0].push(0);
  v[0][0]++;
  for(int i = 0; i < 2 * N; i++) {
    while(!que[i].empty()) {
      int p = que[i].front();
      if(p == N - 1) {
        cout << i << endl;
        return (0);
      }
      for(auto to : graph[p]) dfs(que[i + 1], to.first, to.second);
      que[i].pop();
    }
  }
  cout << -1 << endl;
}