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;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<pair<int, int>>> g(n);
  while (m--) {
    int x, y, z;
    cin >> x >> y >> z;
    x--; y--;
    g[x].emplace_back(y, z);
    g[y].emplace_back(x, z);
  }
  vector<int> d(n, INT_MAX);
  vector<set<int>> s(n);
  priority_queue<pair<int, int>> pq;
  // priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
  pq.emplace(d[0] = 0, 0);
  while (!pq.empty()) {
    int expected, v;
    tie(expected, v) = pq.top();
    pq.pop();
    if (-expected != d[v]) {
      continue;
    }
    // if (expected != d[v]) {
    //   continue;
    // }
    for (const auto& e : g[v]) {
      int to, c;
      tie(to, c) = e;
      int dist = d[v] + !s[v].count(c);
      if (dist < d[to]) {
        s[to].clear();
        s[to].insert(c);
        pq.emplace(-(d[to] = dist), to);
        // pq.emplace(d[to] = dist, to);
      } else if (dist == d[to]) {
        s[to].insert(c);
      }
    }
  }
  cout << (d[n - 1] == INT_MAX ? -1 : d[n - 1]) << '\n';
  return 0;
}