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: Time Limit Exceeded)

#include<iostream>
#include<cstdio>
#include<vector>
#include<set>
#include<algorithm>
#include<unordered_map>
#include<queue>
using namespace std;

int N, M;
typedef pair<int, int> vert;
struct pair_hash{
    size_t operator() (const pair<int, int> &p) const{
        return p.first ^ p.second;
    }
};


int main(){
    cin >> N >> M;
    int p, q, c;
    unordered_map<vert, set<pair<vert, int> >, pair_hash > edges;
    unordered_map<vert, int, pair_hash> d;
    for(int i = 1; i <= N; i++){
        d[make_pair(i, -1)] = 100000000;
    }
    for(int m = 0; m < M; m++){
        scanf("%d %d %d", &p, &q, &c);
        edges[make_pair(p, c)].insert(make_pair(make_pair(q, c), 0));
        edges[make_pair(q, c)].insert(make_pair(make_pair(p, c), 0));
        edges[make_pair(p, -1)].insert(make_pair(make_pair(p, c), 1));
        edges[make_pair(q, -1)].insert(make_pair(make_pair(q, c), 1));
        edges[make_pair(p, c)].insert(make_pair(make_pair(p, -1), 0));
        edges[make_pair(q, c)].insert(make_pair(make_pair(q, -1), 0));
        d[make_pair(p, c)] = d[make_pair(q, c)] = 100000000;
    }

    priority_queue<pair<int, vert> > PQ;
    unordered_map<vert, bool, pair_hash> visited;
    d[make_pair(1, -1)] = 0;
    PQ.push(make_pair(0 * (-1), make_pair(1, -1)));

    while(!PQ.empty()){
        pair<int, vert> now_data = PQ.top(); PQ.pop();
        vert now = now_data.second;
        visited[now] = true;

        if(d[now] < now_data.first * (-1)){
            continue;
        }

        for(set<pair<vert, int> >::iterator itr = edges[now].begin(); itr != edges[now].end(); itr++){
            vert next = (*itr).first;
            if(visited[next]){
                continue;
            }
            if(d[next] > d[now] + (*itr).second){
                d[next] = d[now] + (*itr).second;
                PQ.push(make_pair(d[next]*(-1), next));
            }
            
        }
    }

    int ans = d[make_pair(N, -1)];
    if(ans == 100000000){
        cout << "-1" << endl;
    }else{
        cout << ans << endl;
    }
}