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;
 
queue < int > que[200001];
unordered_map< int, bool > v[100000];
bool visited[100000];
int N, M;
unordered_map< int, vector< int > > g[100001];
vector< pair< int, int > > graph[100000];
 
void dfs(queue< int >& q, int idx, const int cc)
{
  if(v[idx][cc]++) return;
  if(!visited[idx]++) q.push(idx);
  for(auto to : g[idx][cc]) dfs(q, to, cc);
}
 
int main()
{
  cin >> N >> M;
  for(int i = 0; i < M; i++) {
    int p, q, c;
    cin >> p >> q >> c;
    --p, --q;
    g[p][c].push_back(q);
    g[q][c].push_back(p);
    graph[p].push_back({q, c});
    graph[q].push_back({p, c});
  }
 
  que[0].push(0);
  visited[0] = true;
  for(int i = 0; !que[i].empty(); i++) {
    while(!que[i].empty()) {
      int p = que[i].front();
      if(p == N - 1) {
        cout << i << endl;
        return (0);
      }
      for(auto to : graph[p]) dfs(que[i + 1], to.first, to.second);
      que[i].pop();
    }
  }
  cout << -1 << endl;
}