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"
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <map>
#include <vector>
#include <math.h>
#include <algorithm>
#include <queue>
#include <set>
#include <tuple>
using namespace std;

#define FOR(i,init,a) for(int i=init; i<a; i++)
#define rep(i,a) FOR(i,0,a)
#define rrep(i,a) for(int i=a; i>=0; i--)
#define rep1(i,a) for(int i=1; i<=a; i++)
#define cout1(a) cout << a << endl;
#define cout2(a,b) cout << a << " " << b << endl;
#define cout3(a,b,c) cout << a << " " << b << " " << c << endl;
#define cout4(a,b,c,d) cout << a << " " << b << " " << c << " " << d << endl;
#define mem(a,n) memset( a, n, sizeof(a))
#define all(a) a.begin(),a.end()
#define chmin(a,b) a=min(a,b);
#define chmax(a,b) a=max(a,b);

typedef long long ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef vector<int> V;
typedef vector<V> VV;
typedef vector<VV> VVV;
const int INF = 1e9;
const int MOD = 1e9+7;
const ll LLINF = 1e18;
static const double pi = 3.141592653589793;

struct edge { int to, cost; }; 
vector<edge> G[500009];
int N,M;
map<pii,int> id;
int cnt=0;

int dijkstra(int s, int t) {
    
    vector<int> d(cnt,INF);
    priority_queue< pii, vector<pii>, greater<pii> > que;   // first:cost second:頂点 cost基準の昇順
    que.push({0,s});
    d[s] = 0;
    
    while(!que.empty()){
        pii p = que.top(); que.pop();
        int v = p.second;
        if(d[v]<p.first) continue;
        rep(i,G[v].size()){
            edge e = G[v][i];
            if(d[e.to]>d[v]+e.cost){
                d[e.to] = d[v]+e.cost;
                que.push({d[e.to],e.to});
            }
        }
    }
    return d[t];
}

int getid(int node,int com){
    pii p={node,com};
    if(id.count(p)) return id[p];
    return id[p]=cnt++;
}

void addedge(int u,int v,int c){
    u=getid(u,c), v=getid(v,c);
    G[u].push_back({v,0});
    G[v].push_back({u,0});
}

void addedge2(int u,int c){
    int v=getid(u,-1);
    u=getid(u,c);
    G[u].push_back({v,0});
    G[v].push_back({u,1});
}

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    
    cin>>N>>M;
    rep(i,M){
        int u,v,c;
        cin>>u>>v>>c;
        addedge(u,v,c);
        addedge2(u,c);
        addedge2(v,c);
    }
    
    int s=getid(1,-1);
    int t=getid(N,-1);
    int ans=dijkstra(s,t);
    if(ans==INF) ans=-1;
    cout1(ans)
}