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 MN 500000
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*10+ch-'0';ch=getchar();}
	return x*f;
}
int n,m,head[MN+5],cnt,num,vis[MN+5],id[MN+5],q[MN+5],top,d[MN+5];
struct edge{int to,next,w;}e[MN*4+5];
vector<pair<int,int> >v[1000005];
inline void ins(int f,int t,int w){e[++cnt]=(edge){t,head[f],w};head[f]=cnt;}
void Update(int x)
{
    for(int i=head[x];i;i=e[i].next)
        if(d[x]+e[i].w<d[e[i].to])
        {
            d[e[i].to]=d[x]+e[i].w;
            if(e[i].w) q[++top]=e[i].to;
            else Update(e[i].to);
        }
}
int main()
{
    n=num=read();m=read();
    for(int i=1;i<=m;++i)
    {
        int x=read(),y=read(),c=read();
        v[c].push_back(make_pair(x,y));
    }
    for(int i=1;i<=1000000;++i)
        for(int j=0;j<v[i].size();++j)
        {
            int x=v[i][j].first,y=v[i][j].second;
            if(vis[x]!=i) vis[x]=i,id[x]=++num,ins(num,x,0),ins(x,num,1);
            if(vis[y]!=i) vis[y]=i,id[y]=++num,ins(num,y,0),ins(y,num,1);
            ins(id[x],id[y],0);ins(id[y],id[x],0);
        }
    memset(d,40,sizeof(d));d[q[top=1]=1]=0;
    for(int i=1;i<=top;++i) Update(q[i]);
    printf("%d\n",d[n]>1e8?-1:d[n]);
	return 0;
}