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;
struct st
{
	int p,b,c;
};
bool operator<(st a,st b)
{
	return a.c>b.c;
}
vector<int>G[100000];
unordered_map<int,int>d[100000];
unordered_map<int,vector<int>>E[100000];
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	for(int i=0;i<m;i++)
	{
		int p,q,c;
		scanf("%d%d%d",&p,&q,&c);
		p--;
		q--;
		E[p][c].push_back(q);
		E[q][c].push_back(p);
		G[p].push_back(c);
		G[q].push_back(c);
	}
	priority_queue<st>que;
	d[0][0]=0;
	que.push({0,0,0});
	while(!que.empty())
	{
		st p=que.top();
		que.pop();
		if(d[p.p][p.b]!=p.c)
			continue;
		if(d[p.p].find(0)==d[p.p].end()||d[p.p][0]>p.c)
		{
			d[p.p][0]=p.c;
			que.push({p.p,0,p.c});
		}
		if(p.b==0)
		{
			for(int v:G[p.p])
			{
				if(d[p.p].find(v)==d[p.p].end()||d[p.p][v]>p.c+1)
				{
					d[p.p][v]=p.c+1;
					que.push({p.p,v,p.c+1});
				}
			}
			continue;
		}
		for(int v:E[p.p][p.b])
		{
			if(d[v].find(p.b)==d[v].end()||d[v][p.b]>p.c)
			{
				d[v][p.b]=p.c;
				que.push({v,p.b,p.c});
			}
		}
	}
	int Min=INT_MAX;
	for(auto p:d[n-1])
		Min=min(Min,p.second);
	if(Min==INT_MAX)
		puts("-1");
	else
		printf("%d\n",Min);
	return 0;
}