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 <algorithm>
#include <string>
#include <queue>
#include <map>
using namespace std;
typedef pair<int,int> P;

int n,m;
vector<int> cc;

map<P, vector<int> > e;

queue<int> que;
int d[111111];

bool used[111111];
void mv(int c,int k,int a){
  if(d[k] < a) return; 

  if(d[k] > a){
    d[k] = a;
    que.push(k);
  }

  used[k] = true;
  P p = make_pair(c,k);
  for(int i = 0; i < (int)e[p].size(); i++){
    if(!used[e[p][i]]){
      mv(c,e[p][i],a);
    }
  }
}

int main(void){
  scanf("%d%d",&n,&m);
  for(int i = 0; i < m; i++){
    int p,q,c;
    scanf("%d%d%d",&p,&q,&c);
    p--;q--;
    e[make_pair(c,p)].push_back(q);
    e[make_pair(c,q)].push_back(p);
    cc.push_back(c);
  }
  sort(cc.begin(),cc.end());
  cc.erase(unique(cc.begin(),cc.end()),cc.end());

  que.push(0);
  for(int i = 0; i < 111111; i++)
    d[i] = 1<<30;
  d[0] = 0;
  while(!que.empty()){
    int k = que.front();
    que.pop();
    for(int i = 0; i < (int)cc.size(); i++){
      memset(used,false,sizeof(used));
      P p = make_pair(cc[i],k);
      for(int j = 0; j < (int)e[p].size(); j++){
	mv(cc[i],e[p][j],d[k]+1);
      }

    }
  }

  if(d[n-1] >= 1<<30){
    puts("-1");
  }else{
    printf("%d\n",d[n-1]);
  }
}