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<iostream>
#include<queue>
#include<map>
#include<string.h>
#include<algorithm>
using namespace std;

/*
最短路拆点，跑最短路；
*/

const int N=1e6+10;
const int INF=0x3f3f3f3f;

map<pair<int,int>,int>mp;

struct asd{
    int to;
    int w;
    int next;
};
int head[N*8],tol,dd;
asd q[N*8];
int num[N*8];
bool vis[N*8];
int dis[N*8];
int n,m;
int que[N];
int s,t;

int spfa()
{
    s=0;
    t=1;
    for(int i=1;i<=dd+1;i++)
    {
        dis[i]=INF;
        vis[i]=0;
        num[i]=0;
    }
    dis[1]=0;
    vis[1]=num[1]=1;
    que[s]=1;
    while(s<t)
    {
        int u=que[s];
        s++;
        vis[u]=0;
        for(int v=head[u];v!=-1;v=q[v].next)
        {
            int i=q[v].to;
            if(dis[i]>q[v].w+dis[u])
            {
                dis[i]=q[v].w+dis[u];
                if(!vis[i])
                {
                    num[i]++;
                    if(num[i]>=dd)
                        return -2;
                    vis[i]=1;
                    que[t]=i;
                    t++;
                }
            }
        }
    }
    if(dis[n]<INF)
        return dis[n]/2;
    else
        return -1;
}
int getid(int x,int y)
{
    if(mp.find({x,y})!=mp.end())
        return mp[{x,y}];
    return mp[{x,y}]=++dd;
}
void init()
{
    dd=n;
    tol=0;
    memset(head,-1,sizeof(head));
}
void add(int a,int b,int c)
{
    q[tol].w=c;
    q[tol].to=b;
    q[tol].next=head[a];
    head[a]=tol++;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        mp.clear();
        for(int i=0;i<m;i++)
        {
            int x,y,c;
            scanf("%d%d%d",&x,&y,&c);
            int xx=getid(x,c);
            int yy=getid(y,c);
            add(xx,yy,0);
            add(yy,xx,0);
            add(x,xx,1);
            add(xx,x,1);
            add(y,yy,1);
            add(yy,y,1);
        }
        printf("%d\n",spfa());
    }

    return 0;
}