Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in CPP):

Solution:
#include <iostream>
#include <utility>
#include <map>
#include <set>
#include <list>
#include <limits>

using namespace std;

using Pos = pair<int64_t, int64_t>; // station, company

int64_t N, M;

map<Pos, list<Pos> > Way;
set<Pos> Path;

int main()
{

  cin >> N >> M;
  for (int64_t i=0; i<M; ++i) {
    int64_t p, q, c;
    cin >> p >> q >> c;

    Way[make_pair(p, c)].push_back(make_pair(q, c));
    Way[make_pair(q, c)].push_back(make_pair(p, c));

    Way[make_pair(p, c)].push_back(make_pair(p, -1));
    Way[make_pair(p, -1)].push_back(make_pair(p, c));

    Way[make_pair(q, c)].push_back(make_pair(q, -1));
    Way[make_pair(q, -1)].push_back(make_pair(q, c));
  }

  list<Pos> updated;
  for (auto it=Way.lower_bound(make_pair(1, 1)); it!=Way.end() && it->first.first == 1; ++it) {
    if (Path.insert(it->first).second)
      updated.push_back(it->first);
  }

  for (uint64_t cost=1; ! updated.empty(); ++cost) {
    list<Pos> nexts;

    for (const auto& p : updated) {

      for (const auto& q : Way[p]) {
        if (q.first == N) {
          cout << cost << endl;
          return 0;
        }

        if (Path.insert(q).second) {
          if (q.second == -1)
            nexts.push_back(q);
          else
            updated.push_back(q);
        }
      }
    }

    swap(updated, nexts);
  }

  cout << -1 << endl;

  return 0;
}