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<bits/stdc++.h>
using namespace std;
#define maxn 1000005
set<int> col[maxn];
struct Edge
{
	int fr,to,val;
	Edge(){}
	Edge(int a,int b,int c):fr(a),to(b),val(c){}
}eg[maxn<<1];
struct Data
{
	int first,second;
	Data(){}
	Data(int a,int b):first(a),second(b){}
	friend inline bool operator < (Data a,Data b)
		{
			return a.first==b.first?a.second<b.second:a.first<b.first;
		}
};
priority_queue<Data> q;
int head[maxn],edgenum,dis[maxn],vis[maxn];
inline void add(int fr,int to,int val)
{
	eg[++edgenum]=Edge(head[fr],to,val);
	head[fr]=edgenum;
}
inline void dijskra(int st=1)
{
	memset(dis,0x3f,sizeof(dis));
	dis[st]=0;
	q.push(Data(dis[st],st));
	while(!q.empty())
	{
		int id=q.top().second;
		q.pop();
		if(vis[id]) continue;
		vis[id]=1;
		for(int i=head[id];i;i=eg[i].fr)
		{
			int to=eg[i].to,val=!col[id].count(eg[i].val);
			if(dis[to]>dis[id]+val)
			{
				dis[to]=dis[id]+val;
				q.push(Data(-dis[to],to));
				col[to].clear();
				col[to].insert(eg[i].val);
			}
			else if(dis[to]==dis[id]+val)
				col[to].insert(eg[i].val);
		}
	}
}
int main()
{
	int n,m,a,b,c;
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;++i)
	{
		scanf("%d%d%d",&a,&b,&c);
		add(a,b,c),add(b,a,c);
	}
	dijskra();
	if(dis[n]==0x3f3f3f3f) puts("-1");
	else printf("%d\n",dis[n]);
	return 0;
}