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 edge{
	int y,nxt,v;
}e[2000010];
set<int>c[2000010];
int n,m,head[2000010],dis[2000010],vis[2000010],tot;
void add(int x,int y,int v){
	e[++tot]=edge{y,head[x],v};
	head[x]=tot;
}
void dijkstra(){
	priority_queue<pair<int,int> >q;
	memset(dis,0x3f3f3f3f,sizeof(dis));
	dis[1]=0; q.push(make_pair(0,1));
	while(q.size()){
		int x=q.top().second; q.pop();
		if(vis[x]) continue; vis[x]=1;
//		cout<<x<<':'<<dis[x]<<endl;
		for(int i=head[x];i;i=e[i].nxt){
			int y=e[i].y,v=c[x].count(e[i].v)?0:1;
//			cout<<"y = "<<y<<endl;
			if(dis[x]+v<dis[y]){
				dis[y]=dis[x]+v;
				q.push(make_pair(-dis[y],y));
				c[y].clear(); c[y].insert(e[i].v);
			}
			else if(dis[x]+v==dis[y]) c[y].insert(e[i].v);
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
		int x,y,v;
		scanf("%d%d%d",&x,&y,&v);
		add(x,y,v); add(y,x,v);
	}
	dijkstra();
//	cout<<0x3f3f3f3f3f<<endl;
	if(dis[n]==0x3f3f3f3f) printf("-1");
	else printf("%d",dis[n]);
	return 0;
}