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<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<map>
using namespace std;
const int maxn = 2e6+100;
int n,m,cnt;
int x,y,z;
vector<int> f[maxn],g[maxn];
map<pair<int,int>,int> P;
int get(int x,int y)
{
	if(P.find(make_pair(x,y))!=P.end()) return P[make_pair(x,y)];
	else return P[make_pair(x,y)]=++cnt;
}
void add(int x,int y,int z)
{
	f[x].push_back(y);f[y].push_back(x);
	g[x].push_back(z);g[y].push_back(z);
}
int dis[maxn];
bool vis[maxn];
void spfa(int src)
{
	memset(dis,63,sizeof(dis));
	queue<int> Q;
	Q.push(1);
	vis[1]=true;
	dis[1]=0;
	while(!Q.empty())
	{
		int u=Q.front();Q.pop();
		for(int i=0;i<f[u].size();i++)
		{
			int v=f[u][i];
			if(dis[v]>dis[u]+g[u][i])
			{
				dis[v]=dis[u]+g[u][i];
				if(!vis[v])
				{
					vis[v]=true;
					Q.push(v);
				}
			}
		}
		vis[u]=false;
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	cnt=n;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&x,&y,&z);
		int id1=get(x,z),id2=get(y,z);
		add(id1,id2,0);add(x,id2,1);
		add(y,id1,1);
	}
	spfa(1);
	if(dis[n]>1e7) puts("-1");
	else cout<<dis[n]/2;
	return 0;
}