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;
typedef long long ll;
int n,m,nun;
map<int,int> ax[1000005];
map<int,int> bx[1000005];
list<int> edg[400005];
list<int> poi[100005];
int res[100005];
set<pair<int,int> > q;

void ins(int x,int y,int j){
    int a,b;
    if(ax[j][x]==0){
        ax[j][x]=++nun;
        a = nun;
    }
    else a=ax[j][x];
    edg[a].push_back(y);
    poi[x].push_back(j);
}

void play(int x,int val,int j){
    if(res[x]==-1){
        res[x]=val;
        q.insert(make_pair(val,x));
    }
    if(ax[j][x]==0)return;
    list<int>::iterator p;
    int a = ax[j][x];
    ax[j][x]=0;
    for(p=edg[a].begin();p!=edg[a].end();p++){
        play(*p,val,j);
    }
}

int main()
{
    scanf("%d%d",&n,&m);
    int i,j,b,c;
    pair<int,int> a;
    for(i=1;i<=m;i++){
        scanf("%d%d%d",&c,&b,&j);
        ins(c,b,j);
        ins(b,c,j);
    }
    memset(res,-1,sizeof(res));
    q.insert(make_pair(0,1));
    res[1]=0;
    while(!q.empty()){
        a = *(q.begin());
        q.erase(q.begin());
        list<int>::iterator p,p2;
        for(p=poi[a.second].begin();p!=poi[a.second].end();p++){
            b = ax[*p][a.second];
            if(b==0)continue;
            for(p2=edg[b].begin();p2!=edg[b].end();p2++){
                play(*p2,a.first+1,*p);
            }
        }
    }
    printf("%d",res[n]);
}