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<vector>
#include<map>
#include<queue>
#include<algorithm>
#define MAXN 600006
#define INF 2100000000
using namespace std;
struct node
{
	int to,len;
	node(int to=0,int len=0):to(to),len(len){}
};
vector<node> road[MAXN];
int n,m,a,b,camp,in[MAXN],d[MAXN],cnt;
map<int,int> ma[MAXN];
void SPFA()
{
	in[1]=1;
	queue<int> q;
	q.push(1);
	for(int i=2;i<=cnt;i++)
		d[i]=INF;
	while(!q.empty())
	{
		int now=q.front();
		q.pop();
		in[now]=0;
		for(int i=0;i<(int)road[now].size();i++)
		{
			int nex=road[now][i].to;
			if(d[nex]>d[now]+road[now][i].len)
			{
				d[nex]=d[now]+road[now][i].len;
				if(!in[nex])
				{
					in[nex]=1;
					q.push(nex);
				}
			}
		}
	}
}
int main()
{
	scanf("%d%d",&n,&m);
	cnt=n;
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&a,&b,&camp);
		if(!ma[a][camp]) 
		{
			ma[a][camp]=++cnt;
			road[cnt].push_back(node(a,1));
			road[a].push_back(node(cnt,1));
		}
		if(!ma[b][camp]) 
		{
			ma[b][camp]=++cnt;
			road[cnt].push_back(node(b,1));
			road[b].push_back(node(cnt,1));
		}
		road[ma[b][camp]].push_back(node(ma[a][camp],0));
		road[ma[a][camp]].push_back(node(ma[b][camp],0));
	}
	SPFA();
	if(d[n]==INF) printf("-1");
	else printf("%d",d[n]/2);
}