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>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;

struct edge{
    int to;
    int company;

    edge(int t, int c){
        to=t;
        company=c;
    }
};

    deque<pair<int,edge>> Q;
    unordered_set<int> checked[100001];

int main(){
    int M,N;
    cin >> N >> M;
    vector<edge> E[100001];
    for(int i=0;i<M;i++){
        int p,q,c;
        cin >> p >> q >> c;
        E[p].push_back(edge(q,c));
        E[q].push_back(edge(p,c));
    }
    ll d[100001];
    d[1]=0;
    for(int i=2;i<=N;i++){
        d[i]=INT64_MAX;
    }

    Q.push_back(make_pair(0,edge(1,-1)));
    while(!Q.empty()){
        auto t=Q.front();
        Q.pop_front();

        if(t.first>=d[N] || t.second.to==N) break;

        for(auto e : E[t.second.to]){
            if(checked[e.to].count(e.company)>0) continue;
            checked[e.to].insert(e.company);
            if(t.second.company==e.company){
                Q.push_front(make_pair(t.first,e));
                d[e.to]=t.first;
            }else{
                Q.push_back(make_pair(t.first+1,e));
                d[e.to]=t.first+1;
            }
        }
    }
    if(d[N]<INT64_MAX){
        cout << d[N] << endl;
    }else{
        cout << -1 << endl;
    }
    
    return 0;
}