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:
#pragma GCC optimize(2) 
#include<bits/stdc++.h>
#define rg register
#define il inline
using namespace std;
typedef long long ll;
const int mod=1e9+7,O=1000010,inf=0x3f3f3f3f;
struct Edge{int to,nt,w;}e[O];
struct Data{
	int dis,u;
	il bool operator<(Data rhs)const
		{return dis==rhs.dis?u<rhs.u:dis<rhs.dis;}
};
il int gi(){
	rg int o=0,fl=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-')ch=getchar();
	if(ch=='-')fl=-1,ch=getchar();
	while(isdigit(ch))o=o*10+ch-'0',ch=getchar();
	return fl*o;
}
bool vis[O];
set<int>S[O];
priority_queue<Data>q;
int n=gi(),m=gi(),head[O],cnt,dis[O];
il void add(int u,int v,int w){
	e[++cnt]=(Edge){v,head[u],w};
	head[u]=cnt;
}
int main(){
	for(int i=1;i<=m;++i){
		int x=gi(),y=gi(),z=gi();
		add(x,y,z);add(y,x,z);
	}
	memset(dis,0x3f,sizeof dis);
	dis[1]=0;
	q.push((Data){0,1});
	while(!q.empty()){
		int now=q.top().u;q.pop();
		if(vis[now])continue;vis[now]=1;
		for(int i=head[now];i;i=e[i].nt){
			int v=e[i].to,w=!S[now].count(e[i].w);
			if(dis[v]>dis[now]+w){
				dis[v]=dis[now]+w;
				q.push((Data){-dis[v],v});
				S[v].clear();
				S[v].insert(e[i].w);
			}
			else if(dis[v]==dis[now]+w)S[v].insert(e[i].w);
		}
	}
	printf("%d\n",dis[n]==inf?-1:dis[n]);
	return 0;
}