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)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<map>
using namespace std;
const int N=2e6+5;
typedef long long ll;
int n,m;
bool vis[N];
vector<int>edge[N];
queue<int>q;
map<int,int>f[N];
int dist[N];
int fa[N];
int find_father(int x){
    if(x==fa[x]) return x;
    return fa[x]=find_father(fa[x]);
}
void merge(int x,int y){
    int f1=find_father(x),f2=find_father(y);
    fa[f1]=f2;
}
void bfs(){
    memset(dist,-1,sizeof dist);
    q.push(1);
    vis[1]=1;
    dist[1]=0;
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=0,sz=edge[u].size();i<sz;i++){
            int v=edge[u][i];
            v=find_father(v);
            if(!vis[v]){
                dist[v]=dist[u]+1;
                q.push(v);
                vis[v]=1;
            }
        }
    }
}
void Read(int &x){
	char ch = getchar();x = 0;
	for (; ch < '0' || ch > '9'; ch = getchar());
	for (; ch >='0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        int cnt=0;
        for(int i=1;i<=n;i++){
            f[i].clear();
            f[i][0]=++cnt;
        }
        for(int i=1;i<N;i++)
            fa[i]=i;
        for(int i=1;i<=m;i++){
            int u,v,w;
            //scanf("%d%d%d",&u,&v,&w);
            Read(u),Read(v),Read(w);
            if(f[u][w]==0){
                f[u][w]=++cnt;
            }
            u=f[u][w];
            u=find_father(u);
            if(f[v][w]==0){
                f[v][w]=++cnt;
            }
            v=f[v][w];
            v=find_father(v);
            merge(u,v);
        }
        for(int i=1;i<=cnt;i++)
            vis[i]=0,edge[i].clear();
        for(int i=1;i<=n;i++)
            for(map<int,int>::iterator it=f[i].begin();it!=f[i].end();it++){
                int j=it->second;
                j=find_father(j);
                edge[i].push_back(j);
                edge[j].push_back(i);
            }
        bfs();
        if(dist[n]==-1)
            puts("-1");
        else
            printf("%d\n",dist[n]/2);
    }
}