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 fi first
#define se second
#define all(x) (x).begin() , (x).end()
#define rall(x) (x).rbegin() , (x).rend()
#define pb push_back
#define len(x) (int)(x).size()
#define ins insert
typedef long long ll;
typedef long double ld;
using namespace std;
const int N = 1e5, inf = 1e9;
map < int , set < int > > g[N];
bool used[N];
int d[N];
deque < int > layer;
set < pair < int , int > > was;
void dfs(int &st , int v , int C){
    if(d[v] > d[st] + 1){
        d[v] = d[st] + 1;
        layer.pb(v);
    }
    was.insert({v , C});
    for(auto u : g[v][C])
        if(was.find({u , C}) == was.end())
        dfs(st , u , C);
}
signed main(){
    ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0);
    int n , m; cin >> n >> m;
    for(int i = 0 ; i < m; ++i){
        int p , q , c;cin >> p >> q >> c;
        --p, --q;
        g[p][c].ins(q);
        g[q][c].ins(p);
    }


    fill(d , d + N , inf);
    d[0] = 0;
    layer.pb(0);
    while(!layer.empty()){
        int head = layer.front();
        layer.pop_front();
        for(auto C : g[head]){
            for(auto U : C.se)
                if(was.find({U , C.fi}) == was.end())
                    dfs(head , U , C.fi);
        }
    }
    if(d[n - 1] == inf)cout << -1;
    else
    cout << d[n - 1] << '\n';
    return 0;
}