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: Time Limit Exceeded)

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <climits>
#include <map>
using namespace std;
const int SIZEN = 100010;
typedef long long LL;
int N,M;
struct Link{
	int to,next,c;
}link[400010];
int head[SIZEN],tot = 0;
int root[SIZEN] = {0};

int find(int x){
	if(x == root[x])return x;
	return root[x] = find(root[x]);
}

struct Node{
	int x,c,d;
	Node(int a,int b,int t){x = a;c = b; d = t;}
	Node(){;}
	bool operator < (const Node &a)const{
		return x < a.x;
	}
};
map<int,int> dis[SIZEN];
inline void read(int &x){
	char ch;x = 0;
	while(ch = getchar(),ch < '!');
	x = ch-48;
	while(ch = getchar(),ch > '!')x = x*10+ch-48;
}

void add(int from,int to,int c){
	tot ++;
	link[tot].to=to;
	link[tot].c = c;
	link[tot].next=head[from];
	head[from] = tot;
}
queue<Node> q;
int SPFA(int x){
	int ret = INT_MAX;
	priority_queue<Node> q;
	q.push(Node(1,0,0));
	dis[1][0] = 0;
	while(!q.empty()){
		Node t = q.top();
		q.pop();
		if(t.x==N)ret = min(ret,t.d);
		if(t.d!=dis[t.x][t.c] || ret <= dis[t.x][t.c])continue;
		for(int i = head[t.x];i;i=link[i].next){
			if(!dis[link[i].to][link[i].c]||t.d + (!(link[i].c == t.c)) < dis[link[i].to][link[i].c]){
				dis[link[i].to][link[i].c] = t.d+(!(link[i].c == t.c));
				q.push(Node(link[i].to,link[i].c,dis[link[i].to][link[i].c]));
			}
		}
	}
	return ret;
}

int main(){
	read(N);read(M);
	for(int i = 1;i <= N;i++)root[i] = i;
	for(int i = 1;i <= M;i++){
		int a,b,c;
		read(a);read(b);read(c);
		add(a,b,c);add(b,a,c);
		int fa = find(a),fb = find(b);
		if(fa== fb)continue;
		else root[fa] = fb;
	}
	if(find(1) != find(N))puts("-1");
	else printf("%d",SPFA(1));
	return 0;
}
/*
4 4
1 2 1
2 3 1
3 4 1
4 1 1


*/