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 P = pair<int, int>;
using PP = pair<int, P>;
const int INF = 2e9;
struct edge { P to; int cost; };
signed main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int n, m; cin >> n >> m;
  map<P, vector<edge> > G;
  map<P, int> d;
  for (int i = 0; i < m; i++) {
    int p, q, c; cin >> p >> q >> c;
    p--; q--;
    G[P(p, 0)].push_back(edge{P(p, c), 1});
    G[P(p, c)].push_back(edge{P(p, 0), 0});
    G[P(q, 0)].push_back(edge{P(q, c), 1});
    G[P(q, c)].push_back(edge{P(q, 0), 0});
    G[P(p, c)].push_back(edge{P(q, c), 0});
    G[P(q, c)].push_back(edge{P(p, c), 0});
    d[P(p, c)] = d[P(q, c)] = INF;
  }
  for (int i = 0; i < n; i++) d[P(i, 0)] = INF;
  priority_queue< PP, vector<PP>, greater<PP> > pq;
  d[P(0, 0)] = 0;
  pq.push(PP(0, P(0, 0)));
  while (not pq.empty()) {
    PP pp = pq.top(); pq.pop();
    P p = pp.second;
    if (d[p] < pp.first) continue;
    for (auto&& i : G[p]) {
      if (d[p] + i.cost >= d[i.to]) continue;
      d[i.to] = d[p] + i.cost;
      pq.push(PP(d[i.to], i.to));
    }
  }
  cout << (d[P(n - 1, 0)] == INF ? -1 : d[P(n - 1, 0)]) << '\n';
  return 0;
}