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 Rust (1.15.1) (judge verdict: Wrong Answer)

use std::io::stdin;
use std::hash::Hash;
use std::collections::BinaryHeap;
use std::collections::HashSet;
use std::collections::HashMap;
use std::collections::BTreeMap;

fn main(){
    let (n, m) = getpair();
    let mut graph: Graph<(i64, i64)> = Graph::new();
    for _ in 0..m {
        let input: Vec<i64> = getvec();
        
        graph.entry((input[0], input[2])).or_insert_with(BTreeMap::new).insert((input[1], input[2]), 0);
        graph.entry((input[1], input[2])).or_insert_with(BTreeMap::new).insert((input[0], input[2]), 0);
        graph.entry((input[0], -1)).or_insert_with(BTreeMap::new).insert((input[0], input[2]), 1);
        graph.entry((input[1], -1)).or_insert_with(BTreeMap::new).insert((input[1], input[2]), 1);
        graph.entry((input[0], input[2])).or_insert_with(BTreeMap::new).insert((input[0], -1), 0);
        graph.entry((input[1], input[2])).or_insert_with(BTreeMap::new).insert((input[1], -1), 0);
    }
    println!("{}", dijkstra(&mut graph, (1, -1), (n as i64, -1)));
}

#[allow(dead_code)]
fn getline() -> String {
    let mut s = String::new();
    match stdin().read_line(&mut s){
        Ok(_) => {s.trim().to_string()}
        Err(_) => String::new()
    }
}

#[allow(dead_code)]
fn getvec<T: std::str::FromStr>() -> Vec<T>{
    let line = getline();
    line.split_whitespace().filter_map(|x| x.parse().ok()).collect()
}

#[allow(dead_code)]
fn getpair() -> (usize, usize){
    let nm = getvec();
    (nm[0], nm[1])
}

type Graph<N> = HashMap<N, BTreeMap<N, u64>>;

fn dijkstra<N>(graph: &mut Graph<N>, start: N, end: N) -> i64 where N: Eq + Ord + Hash + Copy {
    let mut heap = BinaryHeap::new();
    let mut visited = HashSet::new();

    heap.push((0, start));

    while let Some((cost, current)) = heap.pop() {
        if visited.contains(&current) { continue; }
        if current == end { return cost as i64; }
        visited.insert(current);
        for (next, path_cost) in graph.entry(current).or_insert_with(BTreeMap::new) {
            if !visited.contains(next) {
                heap.push((cost + *path_cost, next.clone()))
            }
        }
    }
    -1
}