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: Wrong Answer)

#include "bits/stdc++.h"
#define ll long long
#define rep2(i,a,b) for(int i=(a);i<=(b);++i)
#define rep(i,n) for(int i=0;i<n;i++)
#define pii pair<int,int>
#define ti3 tuple<int,int,int>
int MOD=1000000000+7;
int INF=1e6;

using namespace std;
string alphabet("abcdefghijklmnopqrstuvwxyz");

main(){
    int n,m,visited[110000]={};
    vector<pii> v[110000];
    cin>>n>>m;
    rep(i,m){
        int a,b,c;
        cin>>a>>b>>c;
        v[a].push_back({b,c});
        v[b].push_back({a,c});
    }
    int num=1;
    queue<int> now;
    now.push(1);
    visited[1]=1;
    while(!now.empty()){
        int t=now.size();
        rep(i,t){
            int a=now.front();
            now.pop();
            rep(j,v[a].size()){
                int d=v[a][j].first;
                int c=v[a][j].second;
                if(visited[d]) continue;
                visited[d]=num;
                now.push(d);
                queue<int> q;
                q.push(d);
                while(!q.empty()){
                    int f=q.front();
                    q.pop();
                    rep(ii,v[f].size()){
                        if(v[f][ii].second==c){
                            if(!visited[v[f][ii].first]) now.push(v[f][ii].first);
                            if(visited[v[f][ii].first]!=num) q.push(v[f][ii].first);
                            visited[v[f][ii].first]=num;
                        }
                    }
                }
            }
        }
        if(visited[n]){
            cout<<num;
            return 0;
        }
        num++;
    }
    cout<<-1;
    return 0;
    
}