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;
inline int read(){
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*f;
}
#define MN 100005
#define inf 200000000
int n,m,d[MN];
struct edge{
	int to,c,nex;
}e[MN*4];int hr[MN],cnt;
inline void ins(int f,int t,int c){
	e[++cnt]=(edge){t,c,hr[f]};hr[f]=cnt;
}
int q[MN*40],top;
bool inq[MN];
set<int> col[MN];
inline void sp(){
	register int i,j;
	for(i=1;i<=n;i++) d[i]=inf;
	inq[1]=1;
	for(d[q[top=i=1]=1]=0;i<=top;inq[q[i++]]=false)
	for(j=hr[q[i]];j;j=e[j].nex){
		bool t=col[q[i]].find(e[j].c)==col[q[i]].end();
		if(d[e[j].to]>d[q[i]]+t){
			if(!inq[e[j].to]) inq[e[j].to]=1,q[++top]=e[j].to;
			col[e[j].to].clear(),col[e[j].to].insert(e[j].c);
			d[e[j].to]=d[q[i]]+t;
		}
		else if(d[e[j].to]==d[q[i]]+t&&col[e[j].to].find(e[j].c)==col[e[j].to].end()){
			if(!inq[e[j].to]) inq[e[j].to]=1,q[++top]=e[j].to;
			col[e[j].to].insert(e[j].c);
		}
	}
}
int main(){
	n=read(),m=read();
	register int i,a,b,c;
	for(i=1;i<=m;i++)
	a=read(),b=read(),c=read(),ins(a,b,c),ins(b,a,c);
	sp();printf(d[n]==inf?"-1":"%d\n",d[n]);
	return 0;
}