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<map>
#include<queue>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 1000000
struct mzls
{
	int to,cost;
}as;
vector<mzls>g[N+5];
inline void insert(int x,int y,int z) 
{
	as.to=y,as.cost=z;
	g[x].push_back(as);
	as.to=x;
	g[y].push_back(as);
}
queue<int>q;
int dist[N+5];
bool f[N+5];
inline void SPFA(int s) 
{
	memset(dist,0x3f3f3f,sizeof(dist));
	q.push(s);
	f[s]=1,dist[s]=0;
	while(!q.empty()) 
	{
		int x=q.front();
		q.pop();
		f[x]=0;
		for(int i=0;i<g[x].size();i++) 
		{
			int v=g[x][i].to;
			if(dist[v]>dist[x]+g[x][i].cost) 
			{
				dist[v]=dist[x]+g[x][i].cost;
				if(!f[v]) 
				{
					f[v]=1;
					q.push(v);
				}
			}
		}
	}
}
map<pair<int,int>,int>fa;
int n,m,t;
int X(int x,int y) 
{
	if(fa[make_pair(x,y)])
		return fa[make_pair(x,y)];
	return fa[make_pair(x,y)]=++t;
}
int main()
{
	scanf("%d%d",&n,&m);
	t=n;
	for(int i=1;i<=m;i++) 
	{
		int x,y,z;
		scanf("%d%d%d",&x,&y,&z);
		int t1=X(x,z);
		int t2=X(y,z);
		insert(t1,t2,0);
		insert(x,t1,1);
		insert(y,t2,1);
	}
	SPFA(1);
	printf("%d\n",dist[n]>=0x3f3f3f?-1:dist[n]/2);
	return 0;
}