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: Accepted)

#include<bits/stdc++.h>
using namespace std;

struct Edge{
    int to;
    int cost;
};
using Graph=vector<vector<Edge>>;
using P = pair<int,int>;

const int INF = 1e9;
int bfs01(int s,int t,const Graph& g){
    int n = g.size();
    vector<int> d(n,INF);
    vector<int> used(n);
    d[s]=0;
    deque<int> deq;
    deq.push_back(s);
    while(!deq.empty()){
        int v = deq.front(); deq.pop_front();
        if(used[v]) continue;
        used[v] = true;
        for(auto &e:g[v]){
            d[e.to] = min(d[e.to],d[v] + e.cost);
            if(e.cost==1) deq.push_back(e.to);
            else deq.push_front(e.to);    
        }
    }
    return d[t];
}
int main(){
    int n,m;
    cin>>n>>m;
    
    Graph g(n+2*m);
    int next_v = n;
    map<P,int> dict;
    auto vertex=[&](int pos,int c){
        if(c==-1) return pos;
        if(!dict.count({pos,c})){
            g[next_v].push_back({pos,0});
            g[pos].push_back({next_v,1});
            dict[{pos,c}]=next_v++;
        }
        return dict[{pos,c}];
    };

    for(int i=0;i<m;i++){
        int p,q,c;
        cin>>p>>q>>c;
        p--,q--,c--;
        int pv = vertex(p,c);
        int qv = vertex(q,c);
        g[pv].push_back({qv,0});
        g[qv].push_back({pv,0});
    }
    
    int ans = bfs01(0,n-1,g);
    if(ans>=INF){
        cout<<-1<<endl;
    }
    else{
        cout<<ans<<endl;
    }

    return 0;
}