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;
#define X first
#define Y second
typedef pair<int,int> P;
const int MAXN=1e5+10,INF=1<<27;
int n,m,x,y,z,dist[MAXN];
set<int> s[MAXN];//±ØÐëÒªÓÃset½øÐÐ´æ´¢£¬ÒòÎª¿ÉÄÜdist×îÐ¡Ê±ÓÐ¶àÖÖ¹«Ë¾¹©Ñ¡Ôñ 
vector<P> G[MAXN];

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++)
	{
		scanf("%d%d%d",&x,&y,&z);
		G[x].push_back(P(y,z));G[y].push_back(P(x,z));
	}
	
	priority_queue<P> q;
	for(int i=2;i<MAXN;i++) dist[i]=INF;
	dist[1]=0;q.push(P(0,1));
	while(!q.empty())
	{
		P t=q.top();q.pop();
		int u=t.Y,cost=-t.X,tmp;
		if(cost>dist[u]) continue;
		for(int i=0;i<G[u].size();i++)
		{
			int v=G[u][i].X,col=G[u][i].Y;
			tmp=dist[u]+(!s[u].count(col));
			if(tmp==dist[v]) s[v].insert(col);
			else if(tmp<dist[v])
			{
				s[v].clear();s[v].insert(col);
				dist[v]=tmp,q.push(P(-dist[v],v));
			}
		}
	}
	printf("%d",(dist[n]==INF)?-1:dist[n]);
	return 0;
}