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>
 
/*
// ordered_set 
# include <ext/pb_ds/assoc_container.hpp>
# include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
# define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
*/
 
using namespace std;
 
typedef long long                                        ll;
typedef long double                                      ld;
typedef pair <int, int>                                  pii;
typedef pair <pii, int>                                  ppi;
typedef pair <int, pii>                                  pip;
typedef pair <pii, pii>                                  ppp;
typedef pair <ll, ll>                                    pll;
 
# define A                                               first
# define B                                               second
# define endl                                            '\n'
# define sep                                             ' '
# define all(x)                                          x.begin(), x.end()
# define kill(x)                                         return cout << x << endl, 0
# define SZ(x)                                           int(x.size())
# define Sort(x)                                         sort(all(x))
# define InTheNameOfGod                                  ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
 
ll power(ll a, ll b, ll md) {return (!b ? 1 : (b & 1 ? a * power(a * a % md, b / 2, md) % md : power(a * a % md, b / 2, md) % md));}
 
const int xn = 1e6 + 10;
const int xm = -20 + 10;
const int SQ = 320;
const int sq = 320 + 10;
const int inf = 1e9 + 10;
const ll INF = 1e18 + 10;
const int MOD = 1e9 + 7;
const int TM = 3;
const int BASE = 257;

int n, m, dist[xn], ans = inf;
vector <pii> adj[xn];
set <int> st[xn], E[xn];
queue <int> q;
bool mark[xn], flag;

void DFS(int v){
	mark[v] = true;
	flag |= (v == n);
	for (pii u : adj[v]) if (!mark[u.A]) DFS(u.A);
}
 
int main(){
    InTheNameOfGod;
    
    cin >> n >> m;
    for (int i = 0; i < m; ++ i){
    	int v, u, c;
    	cin >> v >> u >> c;
    	adj[v].push_back({u, c});
    	adj[u].push_back({v, c});
    	E[c].insert(v);
    	E[c].insert(u);
    	st[v].insert(c);
    	st[u].insert(c);
	}
	st[1].insert(0);
	E[0].insert(1);
	DFS(1);
	if (!flag) kill(-1);
	for (int i = 1; i <= 1e6; ++ i) dist[i] = -1;
	q.push(0);
	while (q.size()){
		int c = q.front();
		q.pop();
		for (int v : E[c]){
			for (int x : st[v]){
				if (dist[x] != -1) continue;
				dist[x] = dist[c] + 1, q.push(x);
			}
			st[v].clear();
		}
	}
	for (pii u : adj[n]) ans = min(ans, dist[u.B]);
	cout << ans << endl;
	
    return 0;
}