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 <cstdio>
#include <iostream>
#include <cmath>
#include <cstring> 
#include <sstream>
#include <algorithm>
#include <cstdlib>
#include <map>
#include <queue>
#include <utility>
#include <vector>
#include <set>
#include <memory.h>
#include <iomanip>
#include <bitset>
#include <list>
#include <stack>
 
using namespace std;
 
#define mod 1000000007

int main()
{
    long long int n, m;
    cin >> n >> m;
    vector<vector<pair<long long int, long long int> > > train(100001);
    for(long long int i = 0; i < m; i++){
        long long int p, q, c;
        cin >> p >> q >> c;
        train[p].push_back(make_pair(q, c));
        train[q].push_back(make_pair(p, c));
    }
    priority_queue<pair<long long int, pair<long long int, long long int> > > qu;
    // cost, nowplace, usedlineの順で
    qu.push(make_pair(0, make_pair(1, 0)));
    vector<long long int> cost(100001, mod);
    cost[1] = 0;
    while(!qu.empty()){
        long long int nowcost = (qu.top()).first;
        long long int nowplace = ((qu.top()).second).first;
        long long int usedline = ((qu.top()).second).second;
        if(nowplace == n){
            cout << nowcost << endl;
            break;
        }
        qu.pop();
        for(long long int i = 0; i < train[nowplace].size(); i++){
            long long int nextplace = train[nowplace][i].first;
            long long int nowline = train[nowplace][i].second;
            long long int nextcost = nowcost;
            if(nowline != usedline) nextcost++;
            if(cost[nextplace] > nextcost){
                cost[nextplace] = nextcost;
                qu.push(make_pair(nextcost, make_pair(nextplace, nowline)));
            }
        }
    }
    if(qu.empty()) cout << -1 << endl;
    return 0;
}