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<cstdio>
#include<set>
#include<queue>
#include<iostream>
#include<cstring>
using namespace std;
struct str{
    int t,n,w;
}e[1000000];
int t,i,h[10000000],u,vv,w,v[10000000],d[10000000],n,m;
set<int>c[1000000];
void add(int u,int v,int w)
{
    t++;
    e[t].t=v;
    e[t].w=w;
    e[t].n=h[u];
    h[u]=t;
}
priority_queue<pair<int,int> >q;
void dij()
{
    memset(d,0x3f,sizeof(d));
    d[1]=0;
    q.push(make_pair(d[1],1));
    while (q.size())
    {
        int u=q.top().second;
        q.pop();
        if (v[u])
            continue;
        v[u]=1;
        for (int i=h[u]; i; i=e[i].n)
        {
            int k=e[i].t,w=c[u].count(e[i].w)?0:1;
            if (d[k]>d[u]+w)
            {
                d[k]=d[u]+w;
                q.push(make_pair(-d[k],k));
                c[k].clear();
                c[k].insert(e[i].w);
            }
            else if (d[k]==d[u]+w)
                c[k].insert(e[i].w);
        }
    }
}
int main()
{
    scanf("%d%d",&n,&m);
    for (i=1; i<=m; i++)
    {
        scanf("%d%d%d",&u,&vv,&w);
        add(u,vv,w);
        add(vv,u,w);
    }
    dij();
    if (d[n]==0x3f3f3f3f)
        printf("-1\n");
    else
        printf("%d\n",d[n]);
}