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<iostream>
#include<vector>
#include<algorithm>
#include<utility>
#include<unordered_map>
#include<queue>
#include<functional>

using namespace std;

using ll = long long int;
using pll = pair<ll, ll>;

int n, m;
struct Edge{
    ll cost, to;
    Edge(ll c, ll t) : cost(c), to(t){}
};
unordered_map<ll, vector<Edge>> g;
unordered_map<ll, ll> d;

ll get_id(ll p, ll c){
    return p + n * c;
}

void add_edge(ll p, ll q, ll c){
    g[get_id(p, c)].push_back({0, get_id(q, c)});
    g[get_id(q, c)].push_back({0, get_id(p, c)});
    g[get_id(p, c)].push_back({0, get_id(p, 0)});
    g[get_id(p, 0)].push_back({1, get_id(p, c)});
    g[get_id(q, c)].push_back({0, get_id(q, 0)});
    g[get_id(q, 0)].push_back({1, get_id(q, c)});
}

int main(){
    cin >> n >> m;
    for(int i = 0; i < m; i++){
        int p, q, c;
        cin >> p >> q >> c;
        add_edge(p, q, c);
    }
    priority_queue<pll, vector<pll>, greater<pll>> q;
    d[get_id(1, 0)] = 0;
    q.push({0, get_id(1, 0)});
    while(!q.empty()){
        auto from = q.top(); q.pop();
        for(auto&& t : g[from.second]){
            if(d.find(t.to) == d.end() || d[t.to] > d[from.second] + t.cost){
                d[t.to] = d[from.second] + t.cost;
                q.push({d[t.to], t.to});
            }
        }
    }
    if(d.find(get_id(n, 0)) == d.end()){
        cout << -1 << endl;
    }else{
        cout << d[get_id(n, 0)] << endl;
    }
    return 0;
}