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<map>
#include<queue>
#include<vector>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
const int Maxn=1e5;
const int INF=0x3f3f3f3f;
struct Edge 
{
    int to,cost;
    Edge(int a,int b){to=a,cost=b;}
};
vector<Edge>way[6*Maxn+5];
void AddEdge(int u,int v,int w) 
{
    way[u].push_back(Edge(v,w));
    way[v].push_back(Edge(u,w));
}
int dis[8*Maxn+5];
int vis[8*Maxn+5];
void spfa(int s) 
{
    memset(dis,0x3f,sizeof dis);
    memset(vis,0,sizeof vis);
    queue<int> q;

    q.push(s);
    vis[s]=1,dis[s]=0;

    while(!q.empty()) 
	{
        int u=q.front();
        q.pop();vis[u]=0;

        for(int i=0;i<way[u].size();i++) 
		{
            int v=way[u][i].to,w=way[u][i].cost;
            if(dis[v]>dis[u]+w) 
			{
                dis[v]=dis[u]+w;
                if(!vis[v]) 
				{
                    vis[v]=1;
                    q.push(v);
                }
            }
        }
    }
}

map<pair<int,int>,int>pos;
int cnt;
int Getid(int u,int c) 
{
    if(pos.find(make_pair(u,c))!=pos.end())
        return pos[make_pair(u,c)];
    return pos[make_pair(u,c)]=++cnt;
}
int t1,t2;
int N,M;
int u,v,c;
int main() 
{
    scanf("%d%d",&N,&M);
    cnt=N;
    for(int i=1;i<=M;i++) 
	{
        scanf("%d%d%d",&u,&v,&c);
        t1=Getid(u,c);
		t2=Getid(v,c);
        AddEdge(t1,t2,0);
        AddEdge(u,t1,1);
        AddEdge(v,t2,1);
    }
    spfa(1);
    if(dis[N]==INF)puts("-1");
    else printf("%d\n",dis[N]/2);
}