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<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<cmath>
#include<cstring>
#include<vector>
#include<climits>
using namespace std;
#define MAXN 400000
#define INF INT_MAX
vector< pair<int,int> > G[MAXN+5];
int n,m,u,v,w;
int dist[MAXN+5];
priority_queue < pair<int,int> > Q;
map<int,int> AC_color[MAXN+5];
void Dij(int s)
{
	Q.push({0,s});
	dist[s]=1;
	while(!Q.empty())
	{
		int x=Q.top().second,Bd=-Q.top().first;
		Q.pop();
		for(int i=0;i<G[x].size();i++)
		{
			int xnext=G[x][i].first;
			int Co=G[x][i].second;
			int NB=Bd;
			if(!AC_color[x].count(Co))NB++;
			if(dist[xnext]>NB)
			{
				AC_color[xnext].clear();
				AC_color[xnext][Co]=1;
				dist[xnext]=NB;
				Q.push({-NB,xnext});
			}
			if(dist[xnext]==NB)AC_color[xnext][Co]=1;
		}
	}
}
int main()
{
	fill(dist+1,dist+MAXN+1,INF);
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&u,&v,&w);
		G[u].push_back({v,w});
		G[v].push_back({u,w});
	}
	Dij(1);
	if(dist[n]!=INF)printf("%d",dist[n]);
	else printf("-1");
	//system("pause");
}