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 <bits/stdc++.h>
 
#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)
 
#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=(int)(a)-1;i>=(int)(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)
 
#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
 
// #define DEBUG
 
#ifdef DEBUG
    #define dump(...) fprintf(stderr, __VA_ARGS__)
#else
    #define dump(...)
#endif
 
template<class T>bool chmax(T &a, const T &b) { return (a<b)?(a=b,1):0;}
template<class T>bool chmin(T &a, const T &b) { return (b<a)?(a=b,1):0;}
 
using namespace std;
using ll=long long;
using vi=vector<int>;
using vll=vector<ll>;
 
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int inf =INT_MAX;
const ll mod=1000000007LL;
const int dx[4]={1,0,-1,0};
const int dy[4]={0,1,0,-1};
 
 
ll extgcd(ll a,ll b,ll& x,ll& y){x=1,y=0;ll g=a;if(b!=0) g=extgcd(b,a%b,y,x),y-=a/b*x;return g;}
ll ADD(const ll &a, const ll &b,const ll &mod) { return (a+b)%mod;}
ll SUB(const ll &a, const ll &b,const ll &mod) { return (a-b+mod)%mod;}
ll MUL(const ll &a, const ll &b,const ll &mod) { return (1LL*a*b)%mod;}
ll DIV(const ll &a, const ll &b,const ll &mod) {ll x,y; extgcd(b,mod,x,y);return MUL(a,(x+mod)%mod,mod);}
 
random_device rd;
mt19937 mt(rd());
uniform_int_distribution<int> dice(1,6);
uniform_real_distribution<double> score(0.0,10.0);

int main(void){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, m; cin >> n >> m;
    using Edge = tuple<int, int>;
    vector<vector<Edge>> G(n);
    rep(loop, m){
        int a, b, c; cin >> a >> b >> c;
        a--, b--;
        G[a].emplace_back(Edge(b, c));
        G[b].emplace_back(Edge(a, c));
    }

    using Elem = tuple<int, int, int>;
    priority_queue<Elem, vector<Elem>, greater<Elem>> q;
    vector<map<int, int>> min_dist(n);
    q.push(Elem(0, 0, 0));
    min_dist[0][0] = 0;

    while(q.size()){
        int cost, v, pc; tie(cost, v, pc) = q.top(); q.pop();
        if(cost != min_dist[v][pc]) continue;

        for(auto& e : G[v]){
            int nv, nc; tie(nv, nc) = e;
            int ncost = cost;
            if(nc != pc) ncost++;
            if(min_dist[nv].find(nc) == end(min_dist[nv]) or ncost < min_dist[nv][nc]){
                min_dist[nv][nc] = ncost;
                q.push(Elem(ncost, nv, nc));
            }
        }
    }

    int res = inf;
    for(auto& e : min_dist[n - 1]){
        chmin(res, e.second);
    }
    if(res == inf) res = -1;
    cout << res << endl;

    return 0;
}