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 <cstdio>
#include <cstring>
#include <iostream>
#include <string>
#include <cmath>
#include <bitset>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <algorithm>
#define INF 1000000000
using namespace std;
typedef long long int ll;
typedef pair<int, int> P;

int main()
{
	int n, m;
	scanf("%d %d", &n, &m);
	if(m==0){
		printf("%d\n", -1);
		return 0;
	}
	vector<P> v[100000];
	for(int i=0; i<m; i++){
		int p, q, c;
		scanf("%d %d %d", &p, &q, &c);
		p--; q--;
		v[p].push_back(P(q, c));
		v[q].push_back(P(p, c));
	}
	int d[100000];
  bool used[100000]={};
	fill(d, d+n, INF);
	d[0]=0;
	priority_queue<P, vector<P>, greater<P> > que;
	que.push(P(0, 0));
	while(!que.empty()){
		P p=que.top();
		que.pop();
		int x=p.second;
      if(used[x]==1) continue;
      used[x]=1;
		for(int i=0; i<v[x].size(); i++){
			int y=v[x][i].first, c=v[x][i].second;
          if(d[y]<d[x]) continue;
			if(d[y]>d[x]+1){
				d[y]=d[x]+1;
				que.push(P(d[y], y));
            }
				vector<int> w;
				set<int> st;
				st.insert(x);
				st.insert(y);
				w.push_back(y);
				while(!w.empty()){
					int z=w.back();
					w.pop_back();
					for(int j=0; j<v[z].size(); j++){
						if(v[z][j].second!=c) continue;
						if(st.find(v[z][j].first)==st.end() && d[v[z][j].first]>=d[x]+1){
                          if(d[v[z][j].first]>d[x]+1){
							d[v[z][j].first]=d[x]+1;
							que.push(P(d[v[z][j].first], v[z][j].first));
                          }
							w.push_back(v[z][j].first);
							st.insert(v[z][j].first);
						}
					}
				}
			
		}
	}
	if(d[n-1]<INF){
		printf("%d\n", d[n-1]);
	}else{
		printf("%d\n", -1);
	}
	return 0;
}