Problem: 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.

Examples

Input

3 3
1 2 1
2 3 1
3 1 2


Output

1


Input

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


Output

2


Input

2 0


Output

-1

Solution (in CPP):

Solution:
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
#include<map>
#include<set>
#define maxn 1000005
#define INF 0x3f3f3f3f
using namespace std;
typedef pair<int,int> PII;
int n,m,cnt,dist[maxn],fa[maxn];
vector<int> G[maxn];
queue<int> q;
map<int,int> id[maxn];
int find(int x)
{
	if(fa[x]==x) return x;
	return fa[x]=find(fa[x]);
}
void bfs()
{
	dist[1]=0;
	q.push(1);
	while(!q.empty())
	{
		int u=q.front(); q.pop();
		for(int i=0;i<G[u].size();i++)
		{
			int v=find(G[u][i]);
			if(dist[v]<0)
			{
				dist[v]=dist[u]+1;
				q.push(v);
			}
		}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<maxn;i++) fa[i]=i;
	for(int i=1;i<=n;i++) id[i][0]=++cnt;
	memset(dist,-1,sizeof(dist));
	for(int i=1;i<=m;i++)
	{
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c);
		if(!id[u][c]) id[u][c]=++cnt;
		u=find(id[u][c]);
		if(!id[v][c]) id[v][c]=++cnt;
		v=find(id[v][c]);
		fa[v]=u;
	}
	for(int i=1;i<=n;i++)
		for(map<int,int>::iterator it=id[i].begin();it!=id[i].end();it++)
		{
			int f1=find(it->second);
			G[i].push_back(f1);
			G[f1].push_back(i);
		}
	bfs();
	printf("%d\n",dist[n]==-1?(-1):(dist[n]/2));
}