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)

//#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast,no-stack-protector") 
//#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>


#define ll long long
#define ull unsigned long long
#define int ll
#define ld long double
#define rep(i, l, r) for (int i = l; i < r; i++)
#define repb(i, r, l) for (int i = r; i > l; i--)
#define sz(a) (int)a.size()
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define pb push_back
#define mp(a, b) make_pair(a, b)
#define ret(a) {cout << a; exit(0);}

using namespace std;

mt19937 mrand(random_device{}()); 
const ll mod=1000000007;
int rnd(int x) { return mrand() % x;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}

typedef pair < int , int > pii ;
typedef pair < long long  , long long > pll ;

const int N = 1e6 + 123, K = 501;
int a[N];
int n , m ;
int c[N] , u[N] , v[N];

vector < pii > gr[N];
set < int > st[N];
int id ;
map < int , int > gid[N];
int d[N];
void flex(int x){
	for(auto to : st[x])
		gid[x][to] = ++id;
}
main(){
	ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
	cin >> n >> m;
	id = n;
	for(int i = 1 ;i <= m; i++){
		cin >> u[i] >> v[i] >> c[i];
		st[u[i]].insert(c[i]);
		st[v[i]].insert(c[i]);
	}
	
	for(int i = 1 ; i <= n; i++)
		flex(i);							
	for(int i = 1; i <= m ; i++){
		int u1 = gid[u[i]][c[i]] , v1 = gid[v[i]][c[i]];
		gr[u1].pb({v1,0});
		gr[v1].pb({u1,0});
		gr[u1].pb({u[i],1});
		gr[v1].pb({v[i],1});
		gr[u[i]].pb({v1,0});
		gr[v[i]].pb({u1,0});
	}
	set < pii > ds;
	for(int i = 1; i <= id; i++)
		d[i] = 1e9 ;
	d[1] = 1;
	ds.insert({d[1] , 1});
	while(sz(ds) != 0){
		int u = (*ds.begin()).se;
		ds.erase(ds.begin());
		for(auto to : gr[u])
			if(d[to.fi] > d[u] + to.se){
				ds.erase({d[to.fi],to.fi});
                d[to.fi] = d[u] + to.se;
				ds.insert({d[to.fi],to.fi});
			}
	}
	int ans = d[n];
	for(auto to : st[n]){
		ans = min(ans , d[gid[n][to]]);
	}

	if(ans >= 1e9)ans = -1;
	cout << ans;

}