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;

struct edge {
  edge() {}
  edge(int to, int cost, int c) : to(to), cost(cost), c(c) {}

  int to, cost, c;
};

int main() {
  int n, m;
  cin >> n >> m;

  map<int, vector<edge>> G[100000];
  vector<map<int, int>> C(100000);
  for (int i = 0; i < m; ++i) {
    int u, v, c;
    cin >> u >> v >> c;
    --u, --v;

    G[u][c].push_back(edge(v, 0, c));
    G[v][c].push_back(edge(u, 0, c));
    C[v][c] = 1;
    C[u][c] = 1;
  }

  for (int i = 0; i < n; ++i) {
    for (auto &v : C[i]) {
      G[i][v.first].push_back(edge(i, 0, 0));
      G[i][0].push_back(edge(i, 1, v.first));
    }
  }

  using P = pair<int, pair<int, int>>;
  const int inf = 1001001001;

  priority_queue<P, vector<P>, greater<P>> pq;
  vector<map<int, int>> d(100000);
  d[0][0] = 0;
  pq.push(P(0, make_pair(0, 0)));

  while (!pq.empty()) {
    P p = pq.top();
    pq.pop();

    int v = p.second.first;
    int c = p.second.second;
    if (d[v][c] < p.first) continue;
    for (auto &e : G[v][c]) {
      if (!d[e.to].count(e.c) || d[e.to][e.c] > d[v][c] + e.cost) {
        d[e.to][e.c] = d[v][c] + e.cost;
        pq.push(P(d[e.to][e.c], make_pair(e.to, e.c)));
      }
    }
  }

  int ans = inf;
  for (auto &i : d[n - 1]) ans = min(ans, i.second);

  if (ans == inf)
    cout << -1 << endl;
  else
    cout << ans << endl;
}