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: Runtime Error)

#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<deque>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned int ui;
const ll mod = 1000000007;
typedef long double ld;
const ll INF = 1e+14;
const int inf = 1e+7;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(int i=sta;i<=n;i++)
typedef complex<ld> Point;
const ld eps = 1e-8;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;
#define fr first
#define sc second

void Yes() {
	cout<<"Yes"<<endl;
	exit(0);
}
 
void No(){
	cout<<"No"<<endl;
	exit(0);
}

struct edge {int to, company;};
int N, M;
vector<edge> G[100100];

int d[100100];

int zero_one_bfs() {
	deque<P> deq;
	rep(i, N) d[i] = inf;
	d[0] = 0;
    deq.push_back(P(0, 0));
	while(deq.size()) {
		P p = deq.front(); deq.pop_front();
		if(p.fr == N - 1) break;
		rep(i, G[p.fr].size()) {
			edge eg = G[p.fr][i];
			if(p.sc == eg.company ) {
				if(d[eg.to] < d[p.fr]) continue;
				deq.push_front(P(eg.to, eg.company));
				d[eg.to] = d[p.fr];
			} else {
				if(d[eg.to] < d[p.fr] + 1) continue;
				deq.push_back(P(eg.to, eg.company));
				d[eg.to] = d[p.fr] + 1;
			}
		}
	}
	if(d[N - 1] == inf) {
		return -1;
	}
	return d[N - 1];
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  rep(i, M) {
		int p, q, c; cin >> p >> q >> c; p --; q --;
		G[p].push_back(edge{q, c});
		G[q].push_back(edge{p, c});
  }
  int res = zero_one_bfs();
  cout << res << "\n";
  return 0;
}