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: Time Limit Exceeded)

#include <iostream>
#include <cstdio>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>

#define rep(i,n) for(int i=0; i<(n); i++)
#define rrep(i,n) for(int i=(n)-1; i>=0; i--)
#define all(X) (X).begin(),(X).end()

#define X first
#define Y second

using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;

template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }

template<class A, size_t N, class T> void Fill(A (&a)[N], const T &v){ fill( (T*)a, (T*)(a+N), v ); }

const int INF = 0x3fffffff;


vector<pii> edge[200005];
vector<pii> v[100005];
int d[200005];
int main(){
	int N, M, ans=0;

	cin >> N >> M;
	rep(i,M){
		int p, q, c;
		cin >> p >> q >> c;
		p--; q--;
		v[p].emplace_back(c,i);
		v[q].emplace_back(c,i);
	}
	int s = M+1;
	int g = M+2;
	v[0].emplace_back( -1, s );
	v[N-1].emplace_back( -1, g );
	rep(i,N){
		rep(j,v[i].size()) rep(k,j){
			int cost = 0;
			if( v[i][j].X != v[i][k].X ){
				cost = 1;
			}
			edge[ v[i][j].Y ].emplace_back( v[i][k].Y, cost );
			edge[ v[i][k].Y ].emplace_back( v[i][j].Y, cost );
		}
	}

	priority_queue<pii, vector<pii>, greater<pii> > que;
	Fill(d, INF);
	d[s] = 0;
	que.emplace(0,s);
	while( !que.empty() ){
		pii p = que.top(); que.pop();
		if( d[p.Y] < p.X ) continue;
		for(auto e: edge[p.Y]){
			if( d[e.X] > d[p.Y] + e.Y ){
				d[e.X] = d[p.Y] + e.Y;
				que.emplace( d[e.X], e.X );
			}
		}
	}

	if( d[g] == INF ) d[g] = 0;

	cout << (d[g]-1) << endl;

	return 0;
}