Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 3
1 2 1
2 3 1
3 1 2

Sample Output 1

1

Use company 1's lines: 1 → 2 → 3. The fare is 1 yen.

Sample Input 2

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

Sample Output 2

2

First, use company 1's lines: 1 → 3 → 2 → 5 → 6. Then, use company 5's lines: 6 → 7 → 8. The fare is 2 yen.

Sample Input 3

2 0

Sample Output 3

-1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
const int NMmax = 100001;
int N,M;
int P[NMmax],Q[NMmax],C[NMmax];
vector<int> ptoc[NMmax];
vector<int> ctop[1000001];

int main(){
    cin >> N >> M;
    for(int i=0; i<M; i++){
        cin >> P[i] >> Q[i] >> C[i];
    }
    for(int i=0; i<M; i++){
        ptoc[P[i]].push_back(C[i]);
        ptoc[Q[i]].push_back(C[i]);
        ctop[C[i]].push_back(P[i]);
        ctop[C[i]].push_back(Q[i]);
    }

    vector<int> cost(NMmax,-1);
    cost[1] = 0;
    queue<int> que;
    que.push(1);
    while(!que.empty()){
        int v = que.front();
        que.pop();
        for(int i=0; i<ptoc[v].size(); i++){
            int c = ptoc[v][i];
            for(int j=0; j<ctop[c].size(); j++){
                int u = ctop[c][j];
                if(cost[u]<0){
                    que.push(u);
                    cost[u]=cost[v]+1;
                }
            }
        }
    }

    cout << cost[N] << endl;
    return 0;
}