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 <iostream>
#include <vector>
#include <unordered_map>
#include <unordered_set>
#include <deque>
int main(){
    int n,m;
    std::cin>>n>>m;
    std::vector<std::unordered_map<int,std::vector<int>>> a(n+1);
    for(int i=0;i<m;i++){
        int p,q,c;
        std::cin>>p>>q>>c;
        a[p][c].push_back(q);
        a[q][c].push_back(p);
    }
    std::deque<std::tuple<int,int,int>> d;
    for(auto e:a[1])
        d.emplace_back(1,1,e.first);
    std::vector<int> b(n+1,n+1);
    std::vector<std::unordered_set<int>> s(n+1);
    while(!d.empty()){
        int v=std::get<0>(d.front());
        int p=std::get<1>(d.front());
        int c=std::get<2>(d.front());
        d.pop_front();
        if(p==n){
            std::cout<<v<<std::endl;
            return 0;
        }
        if(v<b[p]){
            for(auto e:a[p]){
                if(e.first==c)
                    continue;
                for(int q:e.second){
                    if(v+1<=b[q]&&!s[q].count(e.first))
                        d.emplace_back(v+1,q,e.first);
                }
            }
        }
        if(v<=b[p]&&s[p].insert(c).second){
            b[p]=v;
            for(int q:a[p][c]){
                if(v<=b[q]&&!s[q].count(c))
                    d.emplace_front(v,q,c);
            }
        }
    }
    std::cout<<-1<<std::endl;
    return 0;
}