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)

/*感谢MYS大佬和他正经的CSDN....qwq....BY:Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#include<queue>
using namespace std;
const int MAXN=(int)1e5;
const int INF=(int)1e9;
int n_pnt,n_edg,cnt;
map<int,int> ID[MAXN+5];
int input[2*MAXN+5][3];
vector<pair<int,int> > lnk[5*MAXN+5];
int dist[5*MAXN+5];
bool vis[5*MAXN+5];
int main()
{
    scanf("%d%d",&n_pnt,&n_edg);
    cnt=n_pnt+1;
    for(int i=0;i<n_edg;i++)
    {
        int u,v,c;scanf("%d%d%d",&u,&v,&c);
        input[i][0]=u;input[i][1]=v;input[i][2]=c;
        if(!ID[u].count(c)) ID[u][c]=cnt++;
        if(!ID[v].count(c)) ID[v][c]=cnt++;
        lnk[ID[u][c]].push_back(make_pair(ID[v][c],0));
        lnk[ID[v][c]].push_back(make_pair(ID[u][c],0));
    }
    for(int i=1;i<=n_pnt;i++)
        for(map<int,int>::iterator it=ID[i].begin();it!=ID[i].end();it++)
        {
            int id=it->second;
            lnk[i].push_back(make_pair(id,1));
            lnk[id].push_back(make_pair(i,1));
        }
    fill(dist,dist+5*MAXN+5,INF);
    dist[1]=0;vis[1]=true;
    queue<int> que;que.push(1);
    while(!que.empty())
    {
        int u=que.front();que.pop();
        for(int i=0;i<lnk[u].size();i++)
        {
            int v=lnk[u][i].first,l=lnk[u][i].second;
            if(dist[v]>dist[u]+l)
            {
                dist[v]=dist[u]+l;
                if(!vis[v])
                    vis[v]=true,que.push(v);
            }
        }
        vis[u]=false;
    }
    if(dist[n_pnt]==INF) printf("%d\n",-1);
    else printf("%d\n",dist[n_pnt]/2);
    return 0;
}