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;
#define all(vec) vec.begin(),vec.end()
typedef long long int ll;
typedef pair<int,int> P;
const ll MOD=1000000009;
const ll INF=1000000010;
const ll LINF=4000000000000000010LL;
const int MAX=310;
const double EPS=1e-9;
int dx[4]={0,1,0,-1};
int dy[4]={1,0,-1,0};
struct edge{int to, cost;};
vector<edge> G[1100010];
int d[1100010];
int a[200010],b[200010],c[200010];
int main(){
    int n,m;cin>>n>>m;
    int ma=0;
    for(int i=0;i<m;i++){
        cin>>a[i]>>b[i]>>c[i];c[i]--;
        ma=max(ma,c[i]);
    }
    for(int i=0;i<m;i++){
        G[ma+a[i]].push_back({c[i],0});
        G[ma+b[i]].push_back({c[i],0});
        G[c[i]].push_back({ma+a[i],1});
        G[c[i]].push_back({ma+b[i],1});
    }
    priority_queue<P,vector<P>,greater<P>> q;
    fill(d,d+ma+n+1,INF);
    d[ma+1]=0;
    q.push(P(0,ma+1));
    while(!q.empty()){
        int v=q.top().second;q.pop();
        for(auto e:G[v]){
            if(d[e.to]>d[v]+e.cost){
                d[e.to]=d[v]+e.cost;
                q.push(P(d[e.to],e.to));
            }
        }
    }
    if(d[ma+n]==INF){
        cout<<-1<<endl;
        return 0;
    }
    cout<<d[ma+n]<<endl;
    return 0;
}