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 <cstring>
#include <algorithm>
#include <vector>
#include <unordered_map>
#include <deque>
#define inf 0x3F3F3F3F
using namespace std;
struct Node
{
	int x,c;
};
deque<Node>q;
unordered_map<int,vector<int> >e[100005];
unordered_map<int,int>d[100005];
vector<int>g[100005];
int main(void)
{
	int i,x,y,c,cost,n,m,ans=inf;
	scanf("%d%d",&n,&m);
	for(i=1;i<=m;++i)
	{
		scanf("%d%d%d",&x,&y,&c);
		e[x][c].push_back(y);
		e[y][c].push_back(x);
		g[x].push_back(c);
		g[y].push_back(c);
	}
	d[0][0]=0;q.push_back((Node){1,0});
	while(!q.empty())
	{
		x=q.front().x;c=q.front().c;cost=d[x][c];q.pop_front();
		if(d[x].find(0)==d[x].end()||d[x][0]>cost)
		{
			d[x][0]=cost;q.push_front((Node){x,0});
		}
		if(c==0)
		{
			for(i=0;i<g[x].size();++i)
			{
				y=g[x][i];
				if(d[x].find(y)==d[x].end()||d[x][y]>cost+1)
				{
					d[x][y]=cost+1;q.push_back((Node){x,y});
				}
			}
			continue;
		}
		for(i=0;i<e[x][c].size();++i)
		{
			y=e[x][c][i];
			if(d[y].find(c)==d[y].end()||d[y][c]>cost)
			{
				d[y][c]=cost;q.push_front((Node){y,c});
			}
		}
	}
	unordered_map<int,int>::iterator j;
	for(j=d[n].begin();j!=d[n].end();++j)ans=min(ans,j->second);
	if(ans!=inf)printf("%d\n",ans);
	else printf("-1\n");
	return 0;
}