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>
#define N 500005
#define M 500005
using namespace std;
inline int read(){
	int ans=0;
	char ch=getchar();
	while(!isdigit(ch))ch=getchar();
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
	return ans;
}
int n,m,first[N],cnt=0,d[N];
bool vis[N];
struct edge{int v,next,c;}e[M<<1];
inline void add(int u,int v,int c){e[++cnt].v=v,e[cnt].next=first[u],e[cnt].c=c,first[u]=cnt;}
struct heap{int u,v;};
inline bool operator<(heap a,heap b){return a.v==b.v?a.u<b.u:a.v<b.v;}
priority_queue<heap>q;
set<int>col[N*10];
inline void dijkstra(){
	for(int i=2;i<=n;++i)d[i]=0x3f3f3f3f,vis[i]=false;
	q.push((heap){1,d[1]});
	while(!q.empty()){
		heap x=q.top();
		q.pop();
		if(vis[x.u])continue;
		vis[x.u]=1;
		for(int i=first[x.u];i;i=e[i].next){
			int v=e[i].v,w=col[x.u].count(e[i].c)?0:1;
			if(d[v]>d[x.u]+w)d[v]=d[x.u]+w,q.push((heap){v,-d[v]}),col[v].clear(),col[v].insert(e[i].c);
			else if(d[v]==d[x.u]+w)col[v].insert(e[i].c);
		}
	}
}
int main(){
	n=read(),m=read();
	for(int i=1;i<=m;++i){
		int u=read(),v=read(),c=read();
		add(u,v,c),add(v,u,c);
	}
	dijkstra();
	printf("%d",d[n]==0x3f3f3f3f?-1:d[n]);
	return 0;
}