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;
 
typedef long long ll;
const int MAX_N = 100030;    // Change as necessary
const ll  MODD = 1000000009; //
 
int A[MAX_N];
 
vector<pair<int,int> > adj[MAX_N];
queue<int> q;
unordered_set<int> done[MAX_N];
 
void f(int v,int c,int d){
  if(0 <= A[v] && A[v] < d) return;
  if(done[v].count(c)) return;
  done[v].insert(c);
  if(A[v] == -1){
    A[v] = d;
    q.push(v);
  }
  
  for(auto& x : adj[v])
    if(x.second == c)
      f(x.first,c,d);
}
 
int main(){
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  
  int n,m;
  cin >> n >> m;
  
  for(int i=0;i<m;i++){
    int u,v,c;
    cin >> u >> v >> c; u--; v--;
    
    adj[u].emplace_back(v,c);
    adj[v].emplace_back(u,c);
  }
  
  for(int i=0;i<n;i++) A[i] = -1;
  
  q.push(0);
  A[0] = 0;
  while(!q.empty()){
    if(A[n-1] != -1) break;
    int v = q.front(); q.pop();
    
    for(auto& x : adj[v])
      f(x.first,x.second,A[v]+1);
  }
  
  cout << A[n-1] << endl;
  
  return 0;
}