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<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;

int read(){
    int x=0,f=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

#define MAXN 2000000
#define PII pair<int,int>
int N,M;
vector<int> G[MAXN+5];
vector<PII> G1[MAXN+5];

inline void AddEdge(int u,int v){
    G[u].push_back(v);
    G[v].push_back(u);
}

bool vis[MAXN+5];
int bfs(int S,int T){
    queue<PII> Q;
    vis[S]=1;
    Q.push(make_pair(S,0));
    while(!Q.empty()){
        PII tmp=Q.front();Q.pop();
        int u=tmp.first,st=tmp.second;
        for(int i=0;i<int(G[u].size());i++){
            int v=G[u][i];
            if(v==T)
                return st+1;
            if(!vis[v]){
                vis[v]=1;
                Q.push(make_pair(v,st+1));
            }
        }
    }
    return -1;
}

void Get(int S,int c,int d){
    queue<int> Q;
    Q.push(S);vis[S]=1;
    while(!Q.empty()){
        int u=Q.front();Q.pop();
        for(int i=0;i<int(G1[u].size());i++){
            int v=G1[u][i].first,tc=G1[u][i].second;
            if(tc!=c||vis[v])
                continue;
            vis[v]=1;
            Q.push(v);
            AddEdge(u,d);
            AddEdge(v,d);
        }
    }
}

void Build(int C){
    int cnt=N+1;
    for(int i=1;i<=C;i++){
        for(int j=1;j<=N;j++)
            if(!vis[j]){
                bool flag=0;
                for(int k=0;k<int(G1[j].size());k++)
                    if(G1[j][k].second==i){
                        flag=1;
                        break;
                    }
                if(flag){
                    Get(j,i,cnt);
                    cnt++;
                }
            }
        memset(vis,0,sizeof vis);
    }
}

int main(){
    //freopen("data.txt","r",stdin);
    int MaxC=0;
    N=read(),M=read();
    for(int i=1;i<=M;i++){
        int u=read(),v=read(),c=read();
        G1[u].push_back(make_pair(v,c));
        G1[v].push_back(make_pair(u,c));
        MaxC=max(MaxC,c);
    }
    Build(MaxC);
    //for(int i=0;i<int(G[9].size());i++)
    //    printf("%d -> %d\n",9,G[9][i]);
    int Ans=bfs(1,N);
    if(Ans==-1) puts("-1");
    else printf("%d",Ans/2);
}