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++ (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define A first
#define B second
#define db double
#define lowbit(p) (p&(-p))
using namespace std;
void read(int &x){
	x=0; char c=getchar(); int p=1;
	for (;c<48;c=getchar())if (c=='-')p=-1;
	for (;c>47;c=getchar())x=(x<<1)+(x<<3)+(c^48);
	x*=p;
}
void read(ll &x){
	x=0; char c=getchar(); int p=1;
	for (;c<48;c=getchar())if (c=='-')p=-1;
	for (;c>47;c=getchar())x=(x<<1)+(x<<3)+(c^48);
	x*=p;
}
void Min(int &x,int y){
	if (x>y)x=y;
}
void Max(int &x,int y){
	if (x<y)x=y;
}
void Min(ll &x,ll y){
	if (x>y)x=y;
}
void Max(ll &x,ll y){
	if (x<y)x=y;
}

map< pii,int >id;
#define M 100005
struct ed{
	int x,l,nx;
}e[M*12];
int nx[3*M],dis[3*M],n,m,ecnt=1;
bool mark[3*M];
priority_queue< pii >Q;
void add(int x,int y,int l){
	e[ecnt]=(ed){y,l,nx[x]};
	nx[x]=ecnt++;
}
void dij(int s,int t){
	memset(dis,63,sizeof(dis));
	Q.push(mp(dis[s]=0,s));
	int x;
	for (;!Q.empty();){
		x=Q.top().B; Q.pop();
		if (mark[x])continue;
		mark[x]=1;
		if (x==t)break;
		for (int i=nx[x];i;i=e[i].nx){
			if (dis[x]+e[i].l<dis[e[i].x]){
				dis[e[i].x]=dis[x]+e[i].l;
				Q.push(mp(-dis[e[i].x],e[i].x));
			}
		}
	}
	if (mark[t]){
		printf("%d\n",dis[t]);
	}
	else {
		printf("-1\n");
	}
}
int tot;
int get(int x,int y){
	if (id[mp(x,y)])return id[mp(x,y)];
	else return id[mp(x,y)]=++tot;
}
int main(){
	int Caseid;
// 	freopen("4.in","r",stdin);
//	read(Caseid);
	read(n); read(m);	
	int i,x,y,c;
	for (i=1;i<=m;i++){
		read(x); read(y); read(c);
		add(get(x,0),get(x,c),1);
		add(get(x,c),get(y,c),0);
		add(get(y,c),get(y,0),0);
		add(get(y,0),get(y,c),1);
		add(get(y,c),get(x,c),0);
		add(get(x,c),get(x,0),0);
	}
	dij(get(1,0),get(n,0));
	return 0;
}