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<bits/stdc++.h>
#define MAXN 1000010 
#define MAXM 1000010 
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;

struct Edge
{
	int from,to,next,com;
	Edge(){}
	Edge(int u,int v,int n,int c):from(u),to(v),next(n),com(c){}
};

struct Node
{
	int d,u;
	bool operator < (const Node &rhs) const
	{
		return d>rhs.d;
	}
};

int n,m;
Edge edges[MAXM];
int head[MAXN],tot,vis[MAXN],dis[MAXN];
vector<int> comp[MAXN],G[MAXN];

void AddEdge(int u,int v,int com)
{
	edges[tot]=Edge(u,v,head[u],com);
	head[u]=tot++;
	edges[tot]=Edge(v,u,head[v],com);
	head[v]=tot++;
}

set<int> dict;

void init()
{
	dict.clear();
	tot=0;
	memset(head,-1,sizeof(head));
	memset(vis,0,sizeof(vis));
	for(int i=0;i<=2*n;++i)
		G[i].clear(),comp[i].clear();
}

void dijkstra(int s)
{
	memset(dis,0x3f,sizeof(dis));
	memset(vis,0,sizeof(vis));
	priority_queue<Node> q;
	dis[s]=0;
	q.push((Node){0,s});
	while(!q.empty())
	{
		Node x=q.top(); q.pop();
		int u=x.u;
		if(vis[u]) continue;
		vis[u]=1;
		int len=G[u].size();
		for(int i=0;i!=len;++i)
		{
			int v=G[u][i];
			if(dis[v]>dis[u]+1)
			{
				dis[v]=dis[u]+1;
				q.push((Node){dis[v],v});
			}
		}
	}
}

int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		init();
		for(int i=0;i!=m;++i)
		{
			int u,v,com;
			scanf("%d%d%d",&u,&v,&com);
			AddEdge(u,v,com);
			comp[com].push_back(u);
			comp[com].push_back(v);
			dict.insert(com);
		}
		int cnt=0;
		for(set<int>::iterator it=dict.begin();it!=dict.end();++it)
		{
			int com=*it; ++cnt;
			int len=comp[com].size();
			for(int i=0;i!=len;++i)
			{
				int u=comp[com][i];
				if(vis[u]==com) continue;
				vis[u]=com;
				G[u].push_back(n+cnt);
				G[n+cnt].push_back(u);
			}
		}
		dijkstra(1);
		int ans;
		if(dis[n]==INF) ans=-1;
		else ans=dis[n]/2;
		printf("%d\n",ans);
	}
}