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;
typedef double db;
typedef pair<int,int> P;
typedef pair<int,P> iP;
typedef pair<P,int> Pi;
#define pb push_back
#define ft first
#define sd second
#define mp make_pair
#define fr(i,n) for(int i=0;i<(n);i++)
#define Fr(i,n) for(int i=0;i++<(n);)
#define ifr(i,n) for(int i=(n)-1;i>=0;i--)
#define iFr(i,n) for(int i=(n);i>0;i--)

int main(){
  int n,m,p,q,c;
  cin>>n>>m;
  map<P,vector<Pi>> v;
  fr(i,m){
    cin>>p>>q>>c;
    v[mp(p,c)].pb(mp(mp(q,c),0));
    v[mp(q,c)].pb(mp(mp(p,c),0));
    v[mp(p,0)].pb(mp(mp(p,c),1));
    v[mp(p,c)].pb(mp(mp(p,0),0));
    v[mp(q,0)].pb(mp(mp(q,c),1));
    v[mp(q,c)].pb(mp(mp(q,0),0));
  }
  priority_queue<iP,vector<iP>,greater<iP>> pq;
  map<P,bool> vis;
  pq.push(mp(0,mp(1,0)));
  while(!pq.empty()){
    iP ip=pq.top();
    pq.pop();
    int cst=ip.ft;
    P p=ip.sd;
    if(p==mp(n,0)){cout<<cst<<endl;return 0;}
    if(vis[p]) continue;
    vis[p]=1;
    fr(i,v[p].size()){
      if(vis[v[p][i].ft]) continue;
      pq.push(mp(cst+v[p][i].sd,v[p][i].ft));
    }
  }
  cout<<-1<<endl;
}