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<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#include<set>
#include<cctype>
#define pa pair<int,int>
#define INF 0x3f3f3f3f
#define inf 0x3f
#define fi first
#define se second
#define mp make_pair
#define ll long long
#define ull unsigned long long
#define pb push_back

using namespace std;

inline ll read()
{
	long long f=1,sum=0;
	char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-1;c=getchar();}
	while (isdigit(c)) {sum=sum*10+c-'0';c=getchar();}
	return sum*f;
}
const int MAXN=100010;
const int MAXM=200010;
struct edge{
	int next,to,val;
}e[MAXM*10];
int head[MAXN],cnt;
void addedge(int u,int v)
{
	e[++cnt].next=head[u];
	e[cnt].to=v;
	head[u]=cnt;
}
vector <pa> g[MAXN];
vector <int> col[MAXN*10];
int dis[MAXN*10],visit[MAXN],last[MAXN*10]; 
queue <int> q;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for (int i=1;i<=m;i++)
	{
		int u,v,opt;
		scanf("%d%d%d",&u,&v,&opt);
		g[u].push_back(mp(opt,v));
		g[v].push_back(mp(opt,u));
		col[opt].push_back(u);
		col[opt].push_back(v);
	}
	for (int i=1;i<=n;i++)
	{
		sort(g[i].begin(),g[i].end());
		last[i]=0;
	}
	int tot=n+1;
	for (int color=1;color<MAXN;color++)
	{
//		if (color%10000==0) cout<<color<<endl;
		if (col[color].size()==0) continue;
		for (int i=0;i<col[color].size();i++)
		{
			int pt=col[color][i];
			if (visit[pt]==color) continue;
			q.push(pt),visit[pt]=color;
			vector <int> vec;
			vec.clear();
			while (!q.empty())
			{
				int x=q.front();
				vec.push_back(x);
				q.pop();
				while (last[x]<(int)g[x].size())
				{
					int ncol=g[x][last[x]].fi;
					if (ncol!=color) break;
					int v=g[x][last[x]++].se;
					if (visit[v]==color) continue;
					visit[v]=color;
					q.push(v);
				}
			}
			vector <int>::iterator it;
			for (it=vec.begin();it!=vec.end();it++)
				addedge(*it,tot),addedge(tot,*it);
			tot++;
		}
	}
	memset(visit,0,sizeof(visit));
	memset(dis,-1,sizeof(dis));
	q.push(1),dis[1]=0;
	while (!q.empty())
	{
		int x=q.front();
		q.pop();
		for (int i=head[x];i;i=e[i].next)
		{
			int v=e[i].to;
			if (visit[v]) continue;
			visit[v]=1;
			dis[v]=dis[x]+1;
			q.push(v);
		}
	}
	if (dis[n]==-1) cout<<-1;
	else cout<<dis[n]/2;
	return 0;
}
/*
3 3
1 2 1
2 3 1
3 1 2
*/