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 <vector>
using namespace std;

int N, M;
vector<vector<int> >G(100010);
vector<vector<int> >GC(100010);
int s, t;
int H[800100];
int Hn=0;
int Hc[800100];
int H2[400100];
int vis[100010];
int val[100010];
vector<vector<int> >G2(100010);

void addHeap(int nd, int c)
{
	Hn ++;
	H[Hn] = nd;
	Hc[Hn] = c;
	int i=Hn, j;
	while(i>1){
		j = i/2;
		if(Hc[j] > c){
			H[i] = H[j];
			Hc[i] = Hc[j];
			i = j;
		}
		else
			break;
	}
	H[i] = nd;
	Hc[i] = c;
}

void delHeap(int &nd, int &c)
{
	nd = H[1];
	c = Hc[1];
	int tmp = H[Hn];
	int tmpc = Hc[Hn];
	Hn --;
	int i = 1, j;
	while(2*i<=Hn){
		j = 2*i;
		if(j+1<=Hn && Hc[j+1]<Hc[j])
			j ++;
		if(Hc[j] < tmpc){
			H[i] = H[j];
			Hc[i] = Hc[j];
			i = j;
		}
		else
			break;
	}
	H[i] = tmp;
	Hc[i] = tmpc;
} 

int exist(int c, int nd)
{
	int i;
	for(i=0; i<G2[nd].size(); i++){
		if(G2[nd][i] == c)
			return 1;
	}
	return 0;
}

void solve()
{
	int i;
	int res = -1;
	s = 0;
	t = N-1;
	Hn = 0;
	addHeap(s, 0);
	while(Hn > 0){
		int nd, c;
		delHeap(nd, c);
		if(nd == t){
			res = c;
			break;
		}
		if(vis[nd] == 1)
			continue;
		vis[nd] = 1;
		for(i=0; i<G[nd].size(); i++){
			int v = G[nd][i];
			if(vis[v] == 0){
				if(val[v] == 0){
					int c2 = c;
					if( !exist(GC[nd][i], nd) ){
						c2 ++;
					}
					val[v] = c2;
					G2[v].push_back(GC[nd][i]);
					addHeap(v, c2);
				}
				else{
					int c2 = c;
					if( !exist(GC[nd][i], nd) ){
						c2 ++;
					}
					if(c2 == val[v]){
						G2[v].push_back(GC[nd][i]);
						//addHeap(v, c2);
					}
					else if(c2 < val[v]){
						G2[v].clear();
						val[v] = c2;
						G2[v].push_back(GC[nd][i]);
						addHeap(v, c2);
					}
				}
			}
		}
	}
	printf("%d\n", res);
}

int main()
{
	//freopen("in.txt", "r", stdin);
	int i;
	scanf("%d%d", &N, &M);
	for(i=0; i<M; i++){
		int p, q, c;
		scanf("%d%d%d", &p, &q, &c);
		p --;
		q --;
		G[p].push_back(q);
		GC[p].push_back(c);
		G[q].push_back(p);
		GC[q].push_back(c);
	}
	solve();
	return 0;
}