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:
/*
map : count()返回被查找元素的个数 有返回1 无返回0  注意map中不存在相同元素 (下标，即first)
		find()返回位置 若没有则返回map.end()
拆点
*/
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
#define MAXN 100005
#define INF 1000000000
int N,M,cnt;
map<int,int> id[2*MAXN];//id[u][c]:u属性为c的虚点编号
vector< pair<int,int> > G[5*MAXN];
int dist[5*MAXN];
bool vis[5*MAXN];
void mg()
{
	for(int i=0;i<M;i++)
	{
		int u,v,c;
		scanf("%d %d %d",&u,&v,&c);
		if(!id[u].count(c)) 
		{
			id[u][c]=++cnt;
			G[u].push_back(make_pair(cnt,1));
			G[cnt].push_back(make_pair(u,1));
		}
		if(!id[v].count(c)) 
		{
			id[v][c]=++cnt;
			G[v].push_back(make_pair(cnt,1));
			G[cnt].push_back(make_pair(v,1));
		}
		G[id[u][c]].push_back(make_pair(id[v][c],0));
		G[id[v][c]].push_back(make_pair(id[u][c],0));
	}
}
void sp()
{
	fill(dist,dist+5*MAXN,INF);
	dist[1]=0;
	vis[1]=1;
	queue<int> que;
	que.push(1);
	while(!que.empty())
	{
		int u=que.front();
		que.pop();
		for(int i=0;i<G[u].size();i++)
		{
			int v=G[u][i].first,l=G[u][i].second;
			if(dist[v]>dist[u]+l)
			{
				dist[v]=dist[u]+l;
				if(!vis[v])
				{
					vis[v]=1;
					que.push(v);
				}
			}
		}
		vis[u]=0;
	}
}
int main()
{
	scanf("%d %d",&N,&M);
	cnt=N;
	mg();
	sp();
	if(dist[N]==INF) printf("-1\n");
	else printf("%d\n",dist[N]/2);
	return 0;
}