Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
//ios::sync_with_stdio(false);
//cin.tie(0);
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<pii,int> ppii;
typedef pair<int,pii> pipi;
typedef pair<ll,ll> pll;
typedef pair<pll,ll> ppll;
typedef pair<ll,pll> plpl;
typedef pair<ll,plpl> plplp;
typedef tuple<ll,ll,ll> tl;
ll mod=1000000007;
ll mod2=998244353;
ll inf=1000000000000000000;
double pi=2*acos(0);
#define rep(i,m,n) for(int i=m;i<n;i++)
#define rrep(i,n,m) for(int i=n;i>=m;i--)
int dh[4]={1,-1,0,0};
int dw[4]={0,0,1,-1};
int ddh[8]={-1,-1,-1,0,0,1,1,1};
int ddw[8]={-1,0,1,-1,1,-1,0,1};
ll lmax(ll a,ll b){
    if(a<b)return b;
    else return a;
}
ll lmin(ll a,ll b){
    if(a<b)return a;
    else return b;
}
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll n,m;cin>>n>>m;
    vector<pll> v[n+1];
    vector<ll> col[n+1];
    rep(i,0,m){
        ll p,q,c;cin>>p>>q>>c;
        v[p].push_back({q,c});
        v[q].push_back(make_pair(p,c));
        col[p].push_back(c);
        col[q].push_back(c);
    }
    unordered_map<ll,ll> mp[n+1];
    unordered_map<ll,bool> used[n+1];
    rep(i,1,n+1){
        rep(j,0,col[i].size()){
            if(i==1){
                mp[1][col[i][j]]=0;
                used[1][col[i][j]]=false;
            }
            else{
                mp[i][col[i][j]]=inf;
                used[i][col[i][j]]=false;
            }
        }
    }
    ll d[n+1];
    fill(d,d+n+1,inf);
    priority_queue<plplp,vector<plplp>,greater<plplp>> q;
    q.push({0,{1,{-1,-1}}});
    d[1]=0;
    
    
    while(q.size()>0){
        plplp p=q.top();
        q.pop();
        ll now=p.second.first;
        ll dis=p.first;
        ll par=p.second.second.first;
        ll pa=p.second.second.second;
        if(now==n){
            cout<<dis<<endl;
            return 0;
        }
        if(used[now][par])continue;
        used[now][par]=true;
        if(d[now]<dis)continue;
        d[now]=min(d[now],dis);
        rep(i,0,v[now].size()){
            ll ne=v[now][i].first;
            ll co=v[now][i].second;
            if(ne==pa)continue;
            if(par!=co){
                if(mp[ne][co]>dis+1){
                    mp[ne][co]=dis+1;
                    if(d[ne]>=dis+1){
                        q.push({dis+1,{ne,{co,now}}});
                        d[ne]=dis+1;
                    }
                }
            }
            else{
                if(mp[ne][co]>dis){
                    mp[ne][co]=dis;
                    if(d[ne]>=dis){
                        q.push({dis,{ne,{co,now}}});
                        d[ne]=dis;
                    }
                }
            }
        }
    }
    if(d[n]==inf)cout<<-1<<endl;
    else cout<<d[n]<<endl;
}