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 pair<int, int> P;

map<int, vector<int> >m[101010];
map<int, int>flg[101010];
int ans[101010];

int main(){

    for(int i = 0; i < 101010; i++)ans[i] = 101010;

    int N, M;
    cin >> N >> M;
    for(int i = 0; i < M; i++){
        int p, q, c;
        cin >> p >> q >> c;
        m[p][c].push_back(q);
        m[q][c].push_back(p);
    }
    priority_queue<P, vector<P>, greater<P> >Q;
    ans[1] = 0;
    Q.push(P(1, 1));
    while(!Q.empty()){
        int p = Q.top().second;
        Q.pop();
        for(auto it = m[p].begin(); it != m[p].end(); it++){
            int c = (*it).first;
            stack<int>S;
            S.push(p);
            flg[p][c] = 1;
            while(!S.empty()){
                int s = S.top();
                S.pop();
                if(ans[s] == 101010){
                    ans[s] = ans[p] + 1;
                    Q.push(P(ans[s], s));
                }
                for(int i = 0; i < m[s][c].size(); i++){
                    int nxt = m[s][c][i];
                    if(flg[nxt][c] == 0){
                        S.push(nxt);
                        flg[nxt][c] = 1;
                    }
                }
            }
        }
    }    
    if(ans[N] == 101010)cout << "-1" << endl;
    else cout << ans[N] << endl;

    return 0;
}