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 <stdio.h>
#include <string.h>
#include <map>
#include <queue>
#define MN 100000
#define MM 200000
int n,m,vn;
int hd[MN*5+5],to[MM*6+5],nxt[MM*6+5],val[MM*6+5];
int dis[MN*5+5];
bool vis[MN*5+5];
void add(int u,int v,int w){
	static int rn=0;
	to[rn]=v,nxt[rn]=hd[u],val[rn]=w;
	hd[u]=rn++;
}
int gid(int u,int c){
	static std::map<int,int> tb[MN+5];
	if(tb[u].find(c)==tb[u].end()){
		tb[u][c]=(++vn);
		add(u,tb[u][c],1);
		add(tb[u][c],u,0);
	}
	return tb[u][c];
}
void bfs01(){
	static std::deque<int> q;
	memset(dis,0x3f,sizeof(dis));
	dis[1]=0; q.push_front(1);
	while(!q.empty()){
		int u=q.front(); q.pop_front();
		if(vis[u]) continue; vis[u]=true;
		for(int i=hd[u];~i;i=nxt[i]){
			if(vis[to[i]]) continue;
			if(dis[u]+val[i]<dis[to[i]]){
				dis[to[i]]=dis[u]+val[i];
				if(val[i]==0)
					q.push_front(to[i]);
				else
					q.push_back(to[i]);
			}
		}
	}
}
int main(){
	memset(hd,0xff,sizeof(hd));
	scanf("%d%d",&n,&m);
	vn=n;
	for(int i=1,u,v,c;i<=m;i++){
		scanf("%d%d%d",&u,&v,&c);
		u=gid(u,c),v=gid(v,c);
		add(u,v,0),add(v,u,0);
	}
	bfs01();
	if(dis[n]==0x3f3f3f3f) puts("-1");
	else printf("%d\n",dis[n]);
}