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: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define EPS 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define LL long long
const int MOD = 1E9+7;
const int N = 2000000+5;
const int dx[] = {0,0,-1,1,-1,-1,1,1};
const int dy[] = {-1,1,0,0,-1,1,-1,1};
using namespace std;
struct Edge{
    int to,next;
    int val;
    Edge(){}
    Edge(int to,int next,int val):to(to),next(next),val(val){}
    bool operator < (const Edge &rhs)const{
        val>rhs.val;
    }
}edge[N*2];
int head[N*2],tot;
bool vis[N*2];
int dis[N];
void addEdge(int from,int to,int val){
    edge[tot].next=head[from];
    edge[tot].to=to;
    edge[tot].val=val;
    head[from]=tot++;
}
int SPFA(int s,int e){
    memset(vis,false,sizeof(vis));
    memset(dis,INF,sizeof(dis));
    dis[s]=0;
    priority_queue<Edge> Q;
    Q.push(Edge(s,0,0));

    while(!Q.empty()){
        Edge temp=Q.top();
        Q.pop();

        if(!vis[temp.to]){
            vis[temp.to]=true;
            for(int i=head[temp.to];i!=-1;i=edge[i].next){
                if(dis[edge[i].to]>dis[temp.to]+edge[i].val){
                    dis[edge[i].to]=dis[temp.to]+edge[i].val;
                    Q.push(Edge(edge[i].to,0,dis[edge[i].to]));
                }
            }
        }
    }
    if(dis[e]==INF)
        return -1;
    return dis[e]/2;
}
map<int,int> mp[N];
int allPoint;
int getPoint(int x,int y){
    if(!mp[x][y]){
        allPoint++;
        mp[x][y]=allPoint;
    }
    return mp[x][y];
}
int main(){
    int n,m;
    scanf("%d%d",&n,&m);

    allPoint=n;
    memset(head,-1,sizeof(head));
    for(int i=1;i<=m;i++){
        int x,y,w;
        scanf("%d%d%d",&x,&y,&w);

        int newX=getPoint(x,w);
        int newY=getPoint(y,w);

        addEdge(x,newX,1);
        addEdge(newX,x,1);
        addEdge(newX,newY,0);
        addEdge(newY,newX,0);
        addEdge(newY,y,1);
        addEdge(y,newY,1);
    }
    int res=SPFA(1,n);
    printf("%d\n",res);
    return 0;
}