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;

// Graph template
using NodeId = pair<int, int>;

template <class... Args> struct Edge {
  NodeId from, to;
  tuple<Args...> values;
  Edge(NodeId from, NodeId to, Args... values)
      : from(from), to(to), values(values...) {}
};
template <class E> auto operator<(const E &e, const E &f) {
  return e.values > f.values;
}

template <class... Args> class Graph {
  using Edges = vector<Edge<Args...>>;
  map<NodeId, Edges> graph_;

public:
  // Graph(int n) : graph_(n) {}
  size_t size() { return graph_.size(); };
  size_t size() const { return graph_.size(); };
  auto &operator[](NodeId x) { return graph_[x]; }
  const auto &operator[](NodeId x) const { return graph_[x]; }
  void AddEdge(NodeId s, NodeId t, Args... v) {
    graph_[s].emplace_back(s, t, v...);
  }
};

int main() {
  int n, m;
  cin >> n >> m;
  Graph<bool> g;
  for (int i = 0; i < m; ++i) {
    int u, v, c;
    cin >> u >> v >> c;
    g.AddEdge({u, c}, {v, c}, 0);
    g.AddEdge({v, c}, {u, c}, 0);
    g.AddEdge({u, 0}, {u, c}, 1);
    g.AddEdge({u, c}, {u, 0}, 0);
    g.AddEdge({v, 0}, {v, c}, 1);
    g.AddEdge({v, c}, {v, 0}, 0);
  }
  map<NodeId, int> used;
  deque<pair<NodeId, int>> dq;
  dq.push_front({{1, 0}, 0});
  while (!dq.empty()) {
    NodeId u;
    int cost;
    tie(u, cost) = dq.front();
    dq.pop_front();
    if (u == NodeId(n, 0)) {
      cout << cost << endl;
      return 0;
    }
    used[u] = 1;
    for (auto e : g[u]) {
      auto v = e.to;
      if (used[v])
        continue;
      auto weight = get<0>(e.values);
      if (weight) {
        dq.push_back({v, cost + 1});
      } else {
        dq.push_front({v, cost});
      }
    }
  }
  cout << -1 << endl;
}