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)

/*Lucky_Glass*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
using namespace std;
int n_pnt,n_edg,cnt;
map<int,int> ID;
const int MAXN=int(3e5);
vector<int> lnk[MAXN+5];
bool vis[MAXN+5];
int main()
{
	scanf("%d%d",&n_pnt,&n_edg);
	cnt=n_pnt+1;
	for(int i=0;i<n_edg;i++)
	{
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c);
		if(!ID[c]) ID[c]=cnt++;
		c=ID[c];
		lnk[u].push_back(c);lnk[c].push_back(u);
		lnk[v].push_back(c);lnk[c].push_back(v);
	}
	queue<pair<int,int> > que;
	que.push(make_pair(1,0));
	while(!que.empty())
	{
		pair<int,int> fro=que.front();que.pop();
		int u=fro.first,stp=fro.second;
		for(int i=0;i<lnk[u].size();i++)
		{
			int v=lnk[u][i],fstp=stp+1;
			if(vis[v]) continue;
			vis[v]=true;
			if(v==n_pnt) {printf("%d\n",fstp/2);return 0;}
			que.push(make_pair(v,fstp));
		}
	}
	printf("-1\n");
	return 0;
}