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 <bits/stdc++.h>
using namespace std;
typedef pair<int, int> ii;
vector <ii> a[100007];
int n, m;
int d[100007],danh[100007];
void df(int u,int du)
{
   //cout<<u<<" "	;
        for (int i=0;i<a[u].size(); i++)
        {
        	int v=a[u][i].first;
            int uv=a[u][i].second;
            if(uv==du) uv=0;
              else uv=1;
            if (d[v]>d[u]+uv)
			{
                d[v]=d[u]+uv;
                danh[v]=1;
                df(v,a[u][i].second);
                danh[u]=0;
            }
        }
}

main()
{
    int p,q,w;
    cin>>n>>m;
    for(int i=1;i<=m;i++)
	{
        cin>>p>>q>>w;
        a[p].push_back(ii(q, w));
        a[q].push_back(ii(p, w));
    }
    for(int i=1; i<=n; i++) a[i].push_back(ii(0,0));
    for(int i=1;i<=n;i++) d[i]=1000000007;
    d[1]=0;
    df(1,0);
    if(d[n]==1000000007) cout<<-1;
    else cout<<d[n];
}