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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

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

#define llong long long 
#define len(x) ((int)x.size())
#define rep(i,n) for (int i = -1; ++ i < n; )
#define rep1(i,n) for (int i = 0; i ++ < n; )
#define rand __rand
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());  // or mt19937_64
template<class T = int> T rand(T range = numeric_limits<T>::max()) { return (T)(rng() % range); }

#define CONCAT_(x, y) x##y/*{{{*/
#define CONCAT(x, y) CONCAT_(x, y)
#ifdef LOCAL_DEBUG   
int __db_level = 0;
bool __db_same_line = false;
#define clog cerr << string(!__db_same_line ? __db_level * 2 : 0, ' ')
struct debug_block {
    function<void()> fn;
    void print_name() { __db_same_line = true; fn(); clog << endl; __db_same_line = false;  }
    debug_block(function<void()> fn_): fn(fn_) { clog << "{ "; print_name(); ++__db_level; }
    ~debug_block() { --__db_level; clog << "} "; print_name(); }
};
#define DB(args...) debug_block CONCAT(dbbl, __LINE__)([=]{ clog << args; })
#define deb(...)  if (1) { (clog << "[" #__VA_ARGS__ "] = [" << __VA_ARGS__) << "]"; if (!__db_same_line) clog << endl; }
#else
#define clog if (0) cerr
#define DB(...)
#define deb(...)
#endif
template<class T>
ostream& operator,(ostream& out, const T& thing) { return out << ", " << thing; }
template<class U, class V>
ostream& operator<<(ostream& out, const pair<U, V>& p) { return (out << "(" << p.first, p.second) << ")"; }
template<class A, class B>
ostream& operator<<(ostream& out, const tuple<A, B>& t) { return (out << "(" << get<0>(t), get<1>(t)) << ")"; }
template<class A, class B, class C>
ostream& operator<<(ostream& out, const tuple<A, B, C>& t) { return (out << "(" << get<0>(t), get<1>(t), get<2>(t)) << ")"; }
template<class T> ostream& operator<<(ostream& out, const vector<T>& container) { 
    out << "{";
    if (len(container)) out << container[0];
    rep1(i, len(container) - 1) out, container[i];
    return out << "}";
}
template<class x> vector<typename x::value_type> $v(const x& a) { return vector<typename x::value_type>(a.begin(), a.end()); }
#define ptrtype(x) typename iterator_traits<x>::value_type
template<class u> vector<ptrtype(u)> $v(u a, u b) { return vector<ptrtype(u)>(a, b); }/*}}}*/
// ACTUAL SOLUTION BELOW ////////////////////////////////////////////////////////////

const int maxn = 101010;
int n, m;
int dist[maxn];
set<int> reached_from[maxn];
vector<pair<int, int>> gr[maxn];

int main(void) {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cin >> n >> m;
    rep(i, m) {
        int u, v, c; cin >> u >> v >> c;
        gr[u].emplace_back(v, c);
        gr[v].emplace_back(u, c);
    }
    
    fill(dist, dist + n + 1, INT_MAX);
    dist[1] = 0;
    deque<int> qu;
    qu.push_back(1);
    
    for (; len(qu);) {
        int u = qu.front(); qu.pop_front();
        for (auto& [v, c]: gr[u]) {
            int d = !reached_from[u].count(c);
            int new_dist = dist[u] + d;
            if (new_dist == dist[v]) {
                reached_from[v].insert(c);
            }
            if (new_dist >= dist[v]) continue;
            dist[v] = new_dist;
            if (d) qu.push_back(v);
            else qu.push_front(v);
        }
    }
    
    if (dist[n] == INT_MAX) dist[n] = -1;
    cout << dist[n];

    return 0;
}

// Remember:
// - Multitest? REFRESHING the data!!!
// - Constrains for each set of data may differs. Should NOT USE the same max constant (maxn)
//   for all of them.
// vim: foldmethod=marker