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: Accepted)

#include <bits/stdc++.h>
using namespace std;

using tpl = tuple<int, int, int>;
typedef pair<int,int> pii;

void fastStream(){cin.tie(0);std::ios_base::sync_with_stdio(0);}

int N, M;
int P[200010];
int Q[200010];
int C[200010];

map<int, vector<int>> G[200010];
bool used[200100];
int costs[200010];
set<int> used_by_col[200010];

void rec(int pos, int col, queue<int> &q, int cost){
    used_by_col[pos].insert(col);
    if(!used[pos]){
        used[pos] = true;
        costs[pos] = cost;
        q.push(pos);
    }
    for(int to : G[pos][col]){
        if(used_by_col[to].count(col)) continue;
        rec(to, col, q, cost);
    }
}

int main(){

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

    used[0] = true;
    costs[0] = 0;
    queue<int> q;
    q.push(0);
    for(; q.size() > 0; ){
        int pos = q.front(); q.pop();
        int cost = costs[pos] + 1;
        for(auto p : G[pos]){
            int col = p.first;
            auto &v = p.second;
            used_by_col[pos].insert(col);
            for(auto npos : v){
                if(used_by_col[npos].count(col)) continue;
                rec(npos, col, q, cost);
            }
        }
        // for(int i= 0; i < N; i++){
        //     if(used[i]){
        //         cout << i << " ";
        //     }
        // }
        // cout << endl;
        // return 0;
    }
    if(used[N - 1]){
        cout << costs[N - 1] << endl;
    }
    else{
        cout << -1 << endl;
    }

    return 0;
}