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;
const int N=1e5+10,M=5e5+10;
int n,m,d[M],cnt;
map<int,int>g[N];
deque<int>Q;
int head[M],to[M<<2],nxt[M<<2],w[M<<2],tot;

inline int rd()
{
	char ch=getchar();int x=0,f=1;
	while(!isdigit(ch)){if(ch=='-') f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+(ch^48);ch=getchar();}
	return x*f;
}

inline int get(int x,int y)
{
	if(g[x].find(y)!=g[x].end()) return g[x][y];
	else return (g[x][y]=++cnt);
}

inline void lk(int u,int v,int cc)
{to[++tot]=v;nxt[tot]=head[u];head[u]=tot;w[tot]=cc;}

int main(){
	int i,j,x0,y0,xc,yc,ix,iy,iz;
	memset(d,0x3f,sizeof(d));
	n=rd();m=rd();
	for(i=1;i<=n;++i) g[i][0]=++cnt;
	for(;m;m--){
	    ix=rd();iy=rd();iz=rd();
	    x0=get(ix,0);y0=get(iy,0);
	    xc=get(ix,iz);yc=get(iy,iz);
	    lk(xc,yc,0);lk(yc,xc,0);
	    lk(x0,yc,1);lk(yc,x0,1);
	    lk(xc,y0,1);lk(y0,xc,1);
	}
	d[1]=0;
	Q.push_back(1);
	for(;!Q.empty();){
		ix=Q.back();Q.pop_back();
		for(i=head[ix];i;i=nxt[i]){
			j=to[i];
			if(d[j]>d[ix]+w[i]){
				d[j]=d[ix]+w[i];
				if(w[i]==1) Q.push_front(j);
				else Q.push_back(j);
			}
		}
	}
	if(d[n]>M) puts("-1");
	else printf("%d\n",d[n]>>1);
}