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 <bits/stdc++.h>

using namespace std;

using int64 = long long;

int main()
{
  int N, M;
  cin >> N >> M;
  map< int, vector< int > > g[100000];
  for(int i = 0; i < M; i++) {
    int p, q, c;
    cin >> p >> q >> c;
    --p, --q;
    g[p][c].emplace_back(q);
    g[q][c].emplace_back(p);
  }

  queue< int > que;
  int v[100000];
  memset(v, -1, sizeof(v));
  v[0] = 0;
  que.emplace(0);

  while(!que.empty()) {
    int p = que.front();
    que.pop();

    if(p == N - 1) {
      cout << v[p] << endl;
      return (0);
    }

    function< void(int, int) > add_dfs = [&](int company, int idx)
    {
      if(v[idx] == -1) {
        v[idx] = v[p] + 1;
        que.emplace(idx);
      }
      if(!g[idx].count(company)) {
        return;
      }
      auto &x = g[idx][company];
      while(x.size()) {
        int to = x.back();
        x.pop_back();
        add_dfs(company, to);
      }
    };

    for(auto &e : g[p]) {
      add_dfs(e.first, p);
    }
  }

  cout << -1 << endl;
}